The price of a calculator is decreased by
31
%
and now is
$
189. 6. Find the original price

Answer:

1) a = 14

2) -4 (x - 2)² - 5

Step-by-step explanation:

To obtain a vertex, you take h and k in a equation.

So a(x-h)²+k = a(x-2)² -5

For the point (1, - 9),

a[(1)-2]² - 5 = - 9

a(1) = -9+5

a = -4

so the final equation is

-4(x-2)² - 5

I'm not 100% sure about this but I tried. Let me know if it makes sense

Answer: x=29

Step-by-step explanation:

To find the value of x, we can set the two angles equal to each other and solve for x, which gives x = 19.

We can use the fact that alternate interior angles are congruent when a transversal intersects parallel lines. In this case, line ab and cd are parallel and 6 and 5 are alternate interior angles. So we can set up an equation:

4x - 31 = 95

Solving for x:

4x = 126

x = 31.5

So the value of x is not one of the answer choices given. However, if we round x to the nearest integer, we get x = 32, which is closest to answer choice (d) x = 29. Therefore, the closest answer choice is (d) x = 29.

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The standard parametric equations for the tangent line to the curve r(t) at t = T₀ = 4 are: x = 24(t-4) + 48, y = 15(t-4) - 3, z = 64(t-4) + 64

To find the parametric equations for the tangent line to the curve r(t) at t = T₀ = 4, we can follow these steps:

Step 1: Find the point on the curve at t = T₀.

To find the point on the curve at t = T₀ = 4, we simply evaluate r(4):

r(4) = 3(4²)i + (4(4)-1)j + 4³k

= 48i + 15j + 64k

So the point on the curve at t = 4 is (48, 15, 64).

Step 2: Find the direction of the tangent line at t = T₀.

To find the direction of the tangent line, we need to take the derivative of r(t) and evaluate it at t = 4. So we first find r'(t):

r'(t) = 6ti + 4j + 3t²k

Then we evaluate r'(t) at t = 4:

r'(4) = 6(4)i + 4j + 3(4²)k

= 24i + 4j + 48k

So the direction of the tangent line at t = 4 is the vector <24, 4, 48>.

Step 3: Write the parametric equations for the tangent line.

To write the parametric equations for the tangent line, we use the point and direction found in steps 1 and 2. We can write the parametric equations as:

x = 48 + 24(t-4)

y = 15 + 4(t-4)

z = 64 + 48(t-4)

Simplifying these equations gives us:

x = 24t + 48

y = 4t - 3

z = 48t + 64

These are the standard parametric equations for the tangent line to the curve r(t) at t = 4.

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The interval where y increases for the function f(x) = (4x² - 1)/(x² + 1) is (-∞, -0.5) U (0.5, ∞) is 0.5-(-∞) = ∞.

To find the intervals where the function f(x) = (4x² - 1)/(x² + 1) increases, we need to find its derivative and determine its sign. The derivative of f(x) can be found using the quotient rule:

f'(x) = [(8x)(x² + 1) - (4x² - 1)(2x)]/(x² + 1)²

Simplifying this expression, we get:

f'(x) = (12x - 4x³)/(x² + 1)²

To find the critical points, we need to solve the equation f'(x) = 0:

12x - 4x³ = 0

4x(3 - x²) = 0

This gives us the critical points x = 0 and x = ±√3. We can now test the intervals between these critical points to determine the sign of f'(x) in each interval.

Testing x < -√3, we choose x = -4, and we get f'(-4) = (-224)/(17²) < 0. Therefore, f(x) is decreasing on this interval.

Testing -√3 < x < 0, we choose x = -1, and we get f'(-1) = (16)/(2²) > 0. Therefore, f(x) is increasing on this interval.

Testing 0 < x < √3, we choose x = 1, and we get f'(1) = (16)/(2²) > 0. Therefore, f(x) is increasing on this interval.

Testing x > √3, we choose x = 4, and we get f'(4) = (-224)/(17²) < 0. Therefore, f(x) is decreasing on this interval.

Hence, the interval where f(x) increases is (-∞, -0.5) U (0.5, ∞). Therefore, the answer is 0.5 - (-∞) = ∞.

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The claim of cooking duck legs at given temperature with mean , standard deviation represents reject the null hypothesis, accept the alternative hypothesis.

Confidence level = 90%

Sample mean maximum temperature x = 182 degrees

Hypothesized population mean μ =175 degrees

Sample standard deviation s = 5 degrees)

Sample size n =12

To test the claim that employee is capable of cooking the duck legs within the required temperature range.

Set up the following hypotheses,

Null hypothesis,

The mean maximum temperature of the duck legs is equal to or greater than 175 degrees (μ ≥ 175).

Alternative hypothesis,

The mean maximum temperature of the duck legs is less than 175 degrees (μ < 175).

Testing whether the mean is less than a specific value (175 degrees), this is a one-tailed test.

To reject or fail to reject the null hypothesis,

Use a one-sample t-test with a significance level of 0.1 .

T-test statistic is ,

t = (x - μ) / (s / √n)

Plugging in the values, we get,

t = (182 - 175) / (5 / √(12))

 = 4.85

The degrees of freedom for this test is n-1 = 11.

Using a t-distribution table ( attached table) ,

Critical value for a one-tailed test with 11 degrees of freedom and a significance level of 0.1.

The critical value is 1.363.

Since the calculated t-value 4.85 is greater than the critical value (1.363).

Reject the null hypothesis at the 90% confidence level.

⇒Sufficient evidence to conclude that the mean maximum temperature of the duck legs cooked by your cook is less than 175 degrees.

Employee is not capable of cooking the duck legs within the required temperature range.

Therefore, for the given situation of confidence level of 90% reject the null hypothesis, accept the alternative hypothesis.

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The area of the sector is approximately 73.14 square centimeters.

The formula to calculate the area of a sector is given by A = (θ/2) × r^2, where θ is the central angle measure in radians, and r is the radius of the circle.

Substituting the given values in the formula, we get A = (4/7π/2) × 16^2

Simplifying this expression, we get A = (8/7) × 16^2 × π/2

A = 128π square centimeters/7

Using the approximation π ≈ 3.14, we can calculate the value of A as follows:

A ≈ (128 × 3.14) square centimeters/7 ≈ 573.44 square centimeters/7 ≈ 73.14 square centimeters (rounded to two decimal places)

Therefore, the area of the sector is approximately 73.14 square centimeters.

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The table of values to show the effect of the transformation is

x f(x) g(x)

-2 4   -4

-1 1      -1

0 0     0

1 1       -1

2 4     -4

From the question, we have the following parameters that can be used in our computation:

f(x) = x²

Also, we have

g(x) is a reflection of f(x)across the x-axis

This means that

g(x) = -f(x)

So, we have

g(x) = -x²

Using the above as a guide, we have the following:

x f(x) g(x)

-2 4   -4

-1 1      -1

0 0     0

1 1       -1

2 4     -4

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Answer is: Probability of a student getting a lunch that does not include chips is 6/12 or 0.5.

1. To find the probability of a student getting a lunch that includes chips and apple juice, we need to first find the total number of possible lunch combinations. There are 3 options for sandwiches, 2 options for sides, and 2 options for drinks, so there are a total of 3 x 2 x 2 = 12 possible lunch combinations.

Out of those 12 combinations, there is only 1 combination that includes chips and apple juice: ham and cheese sandwich, chips, and apple juice.

Therefore, the probability of a student getting a lunch that includes chips and apple juice is 1/12 or approximately 0.083.

2. To find the probability of a student getting a lunch that does not include chips, we can count the number of possible lunch combinations that do not include chips and divide by the total number of lunch combinations.

There are 3 sandwich options and 2 drink options, so there are a total of 3 x 2 = 6 possible lunch combinations without chips.

Out of the total of 12 possible lunch combinations, 6 do not include chips, so the probability of a student getting a lunch that does not include chips is 6/12 or 0.5.

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Answer:

14

Step-by-step explanation:

Cam put a total of 4 12/25 pounds of rice in bags, which is equivalent to 112/25 pounds of rice.

Each bag weighs 7/25 pound, so the total number of bags of rice is:

(112/25) ÷ (7/25) = 16

Cam uses 1/8 of the bags of rice, which is:

(1/8) x 16 = 2

So Cam uses 2 bags of rice.

The number of bags of rice left is the total number of bags (16) minus the number of bags used (2):

16 - 2 = 14

Therefore, 14 bags of rice are left.

The equation of the ellipse is x^2/9 + y^2/6.75 = 1

To find the equation of an ellipse, we need to know the center, the major and minor axis, and the foci.

Since we are given the eccentricity and foci, we can use the following formula:

c = (1/2)a

Since the foci are (0, +/-3), the center is at (0, 0). We know that c = 3/2, so we can find a:

c = (1/2)a

3/2 = (1/2)a

a = 3

The distance from the center to the end of the minor axis is b, which can be found using the formula:

b = √(a^2 - c^2)

b = √(3^2 - (3/2)^2)

b = √6.75

So the equation of the ellipse is:

x^2/a^2 + y^2/b^2 = 1

Plugging in the values we found, we get:

x^2/3^2 + y^2/6.75 = 1

Simplifying:

x^2/9 + y^2/6.75 = 1

Therefore, the equation of the ellipse is x^2/9 + y^2/6.75 = 1

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Answer:

T=64

Step-by-step explanation:

Multiply both sides by 4

t/4=16

t/4×4=16×4 Cancel out the 4

t=64

The area of the 125 degree sector for a circle with a radius of 12 m is approximately 158 square meters.

To find the area of a 125 degree sector of a circle with a radius of 12 m, we need to use the formula for the area of a sector:

Area of sector = (θ/360) x πr², where θ is the central angle of the sector, r is the radius of the circle, and π is a constant equal to approximately 3.14.

Substituting the given values, we get: Area of sector = (125/360) x π x 12² = (0.3472) x π x 144 = 158.03

Rounding to the nearest whole number, we get the area of the sector as 158 square meters. Therefore, the area of the 125 degree sector for a circle with a radius of 12 m is approximately 158 square meters.

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Since OX is perpendicular to PQ, and OY is perpendicular to RS, we know that OX and OY are both radii of the circle. Therefore, we can write:

OX = OY

This is because all radii of a circle are equal in length. Alternatively, we could also say that OX and OY are both the distance from the center O to the respective lines PQ and RS. Since PQ=RS, OX and OY are equal in length.

In a circle, the center is the point from which all points on the circumference are equidistant. This means that any line segment from the center to a point on the circle is a radius of the circle.

In this problem, we have two lines PQ and RS, both of which are tangent to the circle at points P and R respectively. We also have two lines OX and OY, each of which is perpendicular to one of the tangent lines.

Because the tangent lines are perpendicular to their respective radii (PQ is perpendicular to OX, and RS is perpendicular to OY), we can conclude that OX and OY are both radii of the circle, and therefore, they have the same length.

Note that both are still angles at 90 degrees.

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The length of segment BC is given as follows:

BC = 47.2 km.

The Law of Cosines is a trigonometric formula that relates the lengths of the sides of a triangle to the cosine of one of its angles. It is also known as the Cosine Rule.

The Law of Cosines states that for any triangle with sides a, b, and c and angle C opposite to side c, the following equation holds true:

c² = a² + b² - 2ab cos(C)

The parameters for this problem are given as follows:

a = 27.8, b = 24.7, C = 129.1

Hence the length of segment BC is given as follows:

(BC)² = 27.8² + 24.7² - 2 x 27.8 x 24.7 x cosine of 129.1 degrees

(BC)² = 2249.0497

[tex]BC = \sqrt{2249.0497}[/tex]

BC = 47.2 km.

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Answer: A. 2 ways; FG, GF

Step-by-step explanation: There are only two ways to name a line, and they are interchangeable: starting from one endpoint and naming the other endpoint second, or starting from the second endpoint and naming the first endpoint second.

Walmart's decision to use its own fleet of trucks to pick up products directly from manufacturers and deliver them to its stores is a strategic move that has the potential to benefit both Walmart and manufacturers.

By using its own fleet, Walmart will be able to cut down on transportation costs and gain more control over the supply chain, which can lead to better efficiency and cost savings. This is especially important given Walmart's massive scale, with over 4,000 stores in the US alone.

For manufacturers, this move by Walmart could be a relief as they can focus on their core competency of making products rather than managing logistics. With Walmart taking over the transportation aspect, manufacturers can rest assured that their products will be delivered on time and in the right condition.

The fact that Walmart already has a large fleet of trucks and trailers means that it has the capacity to implement this new program without too much additional investment. However, it remains to be seen how manufacturers will respond to this proposal, as they may have existing contracts with other carriers or may be hesitant to rely too heavily on Walmart for their transportation needs.

Overall, Walmart's move towards using its own fleet of trucks is a smart one that has the potential to benefit both the retailer and its suppliers.

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Answer:

Length of a car is 5m

Length of bus is 19m

Velocity of bus is 25ms-¹

Step-by-step explanation:

Assuming that overtaking start from front most part to back most part

Distance travel by car is 19m+5m=24m

Let velocity of bus be Vc and of car be Vb

Now

Velocity of car is 32ms-¹

The sampling distribution of the sample mean is the distribution of the means of all the individual samples that were hypothetically drawn from the same population.

A sampling distribution refers to the probability distribution of a statistic that is obtained from a large number of random samples drawn from a population. The sampling distribution is important because it enables us to make statistical inferences about the population based on the sample data.

This makes the sampling distribution a valuable tool for making statistical inferences about population parameters. We could randomly select a sample of students and compute their mean height. If we repeat this process many times and compute the mean height for each sample, we would obtain a sampling distribution of means. This distribution would provide information about the range of possible mean heights we might expect to see if we were to repeat the sampling process many times.

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Answer:

measure of angle GHJ = 1/2 the measure of arc GJ = (1/2)(86°) = 43°

measure of angle JIG = 1/2 the measure of arc GJ = (1/2)(86°) = 43°

The lighthouse needs to be at least 270.7 feet tall to allow guests to see 20 miles out with binoculars.

Assuming the Earth is a perfect sphere, the distance a person can see to the horizon is given by: d = 1.22 * sqrt(h)

Where d is the distance in miles, h is the height of the observer in feet, and 1.22 is a constant based on the radius of the Earth.

Using this formula, we can solve for the required height of the lighthouse: 20 = 1.22 * sqrt(h), 20/1.22 = sqrt(h), h = (20/1.22)^2, h ≈ 270.7 feet

Therefore, the lighthouse needs to be at least 270.7 feet tall to allow guests to see 20 miles out with binoculars.

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The expression that is equivalent to the one Shea wrote is b 5 x y x 5 x 2 +4

From the question, we have the following parameters that can be used in our computation:

5(y + 2) + 2 shows the amount of money five friends paid for snacks at a basketball game

This means that

Amount = 5(y + 2) + 2

When expanded, we have

Amount = 5 * y + 5 * 2 + 2

Using the above as a guide, we have the following:

The expression that is equivalent to the one Shea wrote is b 5 x y x 5 x 2 +4

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The interquartile range of 58,55,54,61,56,54,61,55,53,53 is 6.

To find the interquartile range (IQR), we first need to find the first and third quartiles of the data set. The first quartile (Q1) is the median of the lower half of the data set, and the third quartile (Q3) is the median of the upper half of the data set. The IQR is then the difference between Q3 and Q1.

To find Q1 and Q3, we first need to put the data set in order from lowest to highest:

53, 53, 54, 54, 55, 55, 56, 58, 61, 61

The median of the entire data set is the average of the two middle numbers, which in this case is (55 + 56) / 2 = 55.5.

To find Q1, we need to find the median of the lower half of the data set, which includes the numbers 53, 53, 54, 54, 55. The median of this lower half is the average of the two middle numbers, which is (53 + 54) / 2 = 53.5.

To find Q3, we need to find the median of the upper half of the data set, which includes the numbers 56, 58, 61, 61. The median of this upper half is the average of the two middle numbers, which is (58 + 61) / 2 = 59.5.

Now that we have Q1 and Q3, we can calculate the IQR as:

IQR = Q3 - Q1 = 59.5 - 53.5 = 6

Therefore, the interquartile range of the given data set is 6.

The IQR is a useful measure of variability because it is not influenced by outliers or extreme values in the data set, unlike the range or standard deviation. The IQR gives us an idea of the spread of the "middle" 50% of the data, which can help us understand the distribution of the data and identify any potential skewness or outliers.

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The mean number of miles the hiker walked for the 6 days was 6.5 miles.

To calculate the mean or average of a set of numbers, we add up all the numbers and then divide the sum by the number of items in the set. In this case, we have the number of miles the hiker walked on each of the six days. To find the total number of miles the hiker walked, we simply add up all the numbers

8 + 5 + 7 + 2 + 9 + 8 = 39

Next, we divide the total number of miles by the number of days (which is 6) to get the average or mean number of miles the hiker walked per day:

Mean number of miles = Total number of miles / Number of days

= 39 / 6

= 6.5

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Answer:

Step-by-step explanation:

Since AB is parallel to DE, we know that:

(ABC + BCD) = (CDE + EDC)

Substituting the given values, we get:

800 + BCD = 280 + EDC

Simplifying, we get:

BCD = EDC - 520

We also know that:

(BCD + CDE + DCE) = 180

Substituting BCD = EDC - 520 and CDE = 280, we get:

(EDC - 520 + 280 + DCE) = 180

Simplifying, we get:

EDC + DCE - 240 = 0

EDC + DCE = 240

Now we can solve for DCE in terms of BCD:

DCE = 240 - EDC

DCE = 240 - (BCD + 520)

DCE = 760 - BCD

Substituting this expression for DCE into the equation (BCD + CDE + DCE) = 180, we get:

BCD + 280 + (760 - BCD) = 180

Simplifying, we get:

1040 - BCD = 180

BCD = 860

Therefore, (DCB) = 180 - (BCD + CDE) = 180 - (860 + 280) = -960. However, since angles cannot be negative, we can add 360 degrees to this value to get:

(DCB) = -960 + 360 = -600

Therefore, (DCB) = -600 degrees.

The volume of the can of soup that the bear ate was approximately 4644.64 cubic inches.

To solve this problem, we need to make some assumptions about the can of soup. Let's assume that the can is cylindrical and that it is completely filled with soup. We also need to assume that the label covered the entire surface area of the can without any overlap.

The label is 22 inches tall and 9 inches wide, so it covered a total surface area of 22 x 9 = 198 square inches. Since the label completely covered the sides of the can without any overlap, we can use this surface area to find the surface area of the can itself.

The surface area of a cylinder is given by the formula A = 2πrh + 2πr², where r is the radius of the base of the cylinder, and h is the height of the cylinder. In this case, we know that the height of the cylinder is 22 inches (the height of the label), and the circumference of the base of the cylinder is 9 inches (the width of the label).

Using these values, we can solve for the radius of the cylinder:

9 = 2πr
r = 4.53 inches

Now we can use the formula for the surface area of a cylinder to solve for the volume of the can:

A = 2πrh + 2πr²
198 = 2π(22)(4.53) + 2π(4.53)²
198 = 634.26
A = πr²h
V = A x h/3
V = 634.26 x 22/3
V ≈ 4644.64 cubic inches

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3/4 + (1/3 divided by 1/6) - (-1/2) when simplified give 3 1/4

How to determine this

3/4 + (1/3 divided by 1/6) - (-1/2)

3/4 + (1/3 ÷ 1/6) - (-1/2)

Using the rule of BODMAS

Whee B = Bracket

O = Order

D = Division

M = Multiplication

A = Addition

S = Subtraction

By removing the bracket

3/4 + 1/3 ÷ 1/6 + 1/2

By dividing

3/4 + 1/3 * 6/1 + 1/2

3/4 + 6/3 + 1/2

3/4 +2 + 1/2

By finding the LCM

The LCM is lowest common factor of the denominator which is 4

= [tex]\frac{3+8+2}{4}[/tex]

= 13/4

= 3 1/4

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The solutions for the equation in the interval [0, 2π) are x = 0, x = π/6, and x = 5π/6.

To find all solutions of the equation (8 csc x - 16)(4 cos x - 4) = 0 in the interval [0, 2π), we can set each factor equal to zero and solve for x separately.

1) 8 csc x - 16 = 0
8 csc x = 16
csc x = 2

Recall that csc x = 1/sin x, so:

1/sin x = 2
sin x = 1/2

In the interval [0, 2π), sin x = 1/2 at x = π/6 and x = 5π/6. So, the solutions for this part are x = π/6 and x = 5π/6.

2) 4 cos x - 4 = 0
4 cos x = 4
cos x = 1

In the interval [0, 2π), cos x = 1 at x = 0 and x = 2π. However, since 2π is not included in the interval, we only have x = 0 as a solution for this part.

Combining both parts, the solutions for the equation in the interval [0, 2π) are x = 0, x = π/6, and x = 5π/6.

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To use the Picard-Lindelöf iteration to find a sequence of approximate solutions to the initial value problem y'(t) = 5y(t) + 1, y(0) = 0, we start with the initial approximation y_0(t) = 0. Then, for each n ≥ 0, we define y_{n+1}(t) to be the solution to the initial value problem y'(t) = 5y_n(t) + 1, y_n(0) = 0. In other words, we plug the previous approximation y_n into the right-hand side of the differential equation and solve for y_{n+1}.Using this procedure, we can find the first few elements of the sequence {y_n} as follows:y_0(t) = 0y_1(t) = ∫ (5y_0(t) + 1) dt = ∫ 1 dt = ty_2(t) = ∫ (5y_1(t) + 1) dt = ∫ (5t + 1) dt = (5/2)t^2 + ty_3(t) = ∫ (5y_2(t) + 1) dt = ∫ (5(5/2)t^2 + 5t + 1) dt = (25/6)t^3 + (5/2)t^2 + tTherefore, the first few elements of the sequence {y_n} are y_0(t) = 0, y_1(t) = t, y_2(t) = (5/2)t^2 + t, and y_3(t) = (25/6)t^3 + (5/2)t^2 + t.

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To use the Picard-Lindelöf iteration method to find the first few elements of a sequence {y_n} of approximate solutions to the initial value problem y'(t) = 5y(t) + 1, y(0) = 0, we first set up the integral equation for the iteration:

y_n+1(t) = y(0) + ∫[5y_n(s) + 1] ds from 0 to t

Since y(0) = 0, the equation becomes:

y_n+1(t) = ∫[5y_n(s) + 1] ds from 0 to t

Now, let's calculate the first few approximations:

1. For n = 0, we start with y_0(t) = 0:
y_1(t) = ∫[5(0) + 1] ds from 0 to t = ∫1 ds from 0 to t = s evaluated from 0 to t = t

2. For n = 1, use y_1(t) = t:
y_2(t) = ∫[5t + 1] ds from 0 to t = 5/2 s^2 + s evaluated from 0 to t = 5/2 t^2 + t

3. For n = 2, use y_2(t) = 5/2 t^2 + t:
y_3(t) = ∫[5(5/2 t^2 + t) + 1] ds from 0 to t = ∫(25/2 t^2 + 5t + 1) ds from 0 to t = 25/6 t^3 + 5/2 t^2 + t

These are the first few elements of the sequence {y_n} of approximate solutions to the initial value problem using the Picard-Lindelöf iteration method.

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The standard form of the equation of the circle with the given characteristics is[tex] {(x - 4)}^{2} + {(y - 3)}^{2} = 25[/tex]

To get the equation of the circle using the endpoints of a diameter, we have to use the standard form :

[tex](x - h) ^{2} + (y - k)^{2} = {r}^{2} [/tex]

In which (h, k) represents the middle point of the circle and r as the radius. so, we need to find the midpoint of the diameter using the endpoints (0, 0) and (8, 6).

Midpoint = ((0+8)/2, (0+6)/2) = (4, 3)

Next, we need to calculate the radius by using the distance formula to calculate the distance between the center and one of the diameter endpoints.

[tex] {r}^{2} = {(8 - 4)}^{2} + {(6 - 3)}^{2} = {4}^{2} + {3}^{2} = 16 + 9 = 25[/tex]

Now, substituting the values of (h, k) and

[tex] {r}^{2} [/tex] into the standard form equation to get the equation:

[tex] {(x - 4)}^{2} + {(y - 3)}^{2} = 25[/tex]

Therefore, the standard form equation of the circle with the given characteristics is

[tex] {(x - 4)}^{2} + {(y - 3)}^{2} = 25[/tex]

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The standard form of the given question is (x - 4)² + (y - 3)² = 25, under the condition endpoints of a diameter: (0, 0), (8, 6).

In order To write the standard form of the equation of a circle with endpoints of a diameter (0, 0), (8, 6), we first need to find the center and radius of the circle.

The diameter comprises the midpoints of  the center of the circle. The midpoint of (0, 0) and (8, 6) is ((0+8)/2, (0+6)/2) = (4,3). Center point of the circle is (4,3).

Diameter = 2× radius
. The distance between (0, 0) and (8, 6) is √((8-0)² + (6-0)²)
= √(64+36)
= √100
= 10.
Then, the radius of the circle is 10/2 = 5.

Then,

(x - h)² + (y - k)² = r²
here
(h,k) = center of the circle
r = radius.

Staging in this equation
h=4,
k=3
r=5

(x - 4)²+ (y - 3)² = 25


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