four customer's purchased 4 donuts that weigh 3.4 pounds, how much did all four donuts weigh

Answer:

3.5 seconds

Step-by-step explanation:

h(t) = 112t - 16t^2

we can re-write it: h(t) = -16t^2 + 112t

with a = -16 and b = 112

the formula -b/(2a) corresponds to x-coordinate of the maximum.

-112/(2 x -16) = 3.5

The number of National League players cards in the card collection in terms of the number of American League players cards can be expressed as 5x.

Let's start with the information given in the problem:

The baseball card collection contains five times as many National League players as American League players.

Let's use x to represent the number of American League players cards in the collection.

Since the collection contains five times as many National League players as American League players, we can express the number of National League players cards in terms of x as follows:

Number of National League players cards = 5 times the number of American League players cards

Number of National League players cards = 5x

So, if the number of American League players cards in the collection is x, then the number of National League players cards would be 5x, according to the problem statement.

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Answer:

h ≈ 19 m

Step-by-step explanation:

using the tangent ratio in the right triangle

tan43° = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{h}{20}[/tex] ( multiply both sides by 20 )

20 × tan43° = h , then

h ≈ 19 m ( to the nearest whole number )

FIFO Method:

1. Determine the cost of the oldest inventory items available for sale (i.e., the items purchased first.

2.Multiply the cost per item by the number of items sold to determine the COGS.

3.Subtract the COGS from the total cost of the oldest inventory items to determine the value of the last inventory.

LIFO Method: The only difference between FIFO and LIFO method is that in LIFO method, newest inventory cost and value are determined in the first and last step respectively.

The average cost method is a way of determining the cost of inventory by taking the average of the cost of each item in the inventory. FIFO and LIFO are two other methods of determining the cost of inventory.

Average Cost Method:

Determine the total cost of all inventory items available for sale.

Determine the total number of inventory items available for sale.

Divide the total cost of inventory by the total number of items to get the average cost per item.

Multiply the average cost per item by the number of items sold to determine the cost of goods sold (COGS).

Subtract the COGS from the total cost of inventory to determine the value of the ending inventory.

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therefore , Kim earn Average is: $11.70(rounded) per hour

Total for two nights:$116.95

Friday, total earning:

(11 - 6)*$5.75 + $27.35

5×$5.75+$27.35

= $56.1

Saturday, total earning:

(11 - 6)*$5.75 + $32.10

5×$5.75+$32.10

= $60.85

Total for the two nights is:

$56.1 + $60.85 = $116.95

Define average per hour?

The term "hourly average" refers to the average result obtained from ongoing monitoring over each 60-minute interval. An arithmetic average of all entire 15-minute data blocks for a clock hour is referred to as the hourly average.

Average per hour is:

$116.95/(5 + 5)

= $116.95 / 10

= $11.70(rounded)

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Using Pythagoras theorem, the value of c is √193.

What is Pythagoras' theorem?

A right triangle's three sides are related in Euclidean geometry by the Pythagorean theorem, also known as Pythagoras' theorem. According to this statement, the areas of the squares on the other two sides add up to the size of the square whose side is the hypotenuse.

Here, we have

Given:  the right triangle and given sides.

We have to find the value of c.

We apply here Pythagoras theorem and we get

a² + b² = c²

a = 7

b = 12

c = ?

(7)² + (12)² = c²

49 + 144 = c²

193 = c²

c = √193

Hence, the value of c is √193.

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The height of the flagpole is approximately 68.53 feet.

Trigonometric equations are equations that involve trigonometric functions such as sine, cosine, tangent, and their reciprocals. These equations typically involve finding the values of the angles or sides of a right-angled triangle or periodic functions that repeat themselves after certain intervals.

This problem involves finding the height of a flagpole using trigonometry. The surveyor measures the angle of elevation from the tripod to the top of the pole and knows the distance from the base of the pole to the tripod. By using the tangent function and the given angle, we can set up an equation that relates the height of the pole to the distance and the angle. We simplify the equation and solve for the unknown height, which gives us an approximate value of 68.53 feet.

the height of the flagpole, we can use trigonometry. Let's call the height of the flagpole "h".

Using the tangent function, we can set up the following equation:

tan(26) = h / (150 + 3)

The 3 feet is added to the distance because the tripod is 3 feet tall.

Simplifying this equation, we get:

tan(26) = h / 153

To solve for "h", we can multiply both sides by 153:

h = tan(26) * 153

Using a calculator, we can evaluate the right side of the equation to get:

h ≈ 68.53

Therefore, the height of the flagpole is approximately 68.53 feet.

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hi i also need help me toppers

Answer:

125cm²

Step-by-step explanation:

The Answer: 758.83 in³

Step-by-step explanation:

Volume of Solids

The volume of the solid is 758.83 in³

Step-by-step explanation:

Given that the cone and half sphere is hollow

The volume of the cone =

The Volume of the sphere  =

So the Volume of the half sphere =

The volume of solid = volume of cone + volume of the half sphere

Given

height h = 19 in

radius r = 5 in

 =

    = 1/3 × 3.14 × 5 × 5 × 19

    = 497.16 in³

  = 2/3 × 3.14 × 5 × 5 × 5

  = 261.67 in³

V =  497.16 in³ + 261.67 in³

 = 758.83 in³

Hence the volume of the solid is 758.83 in³

Volume of Solids

The volume of the solid is 2813 in³

Step-by-step explanation:

Given that the cone and half sphere is hollow

The volume of the cone  [tex]=1/3\pi r^2h[/tex]

The Volume of the sphere [tex]=4/3\pi r^3[/tex]

So the Volume of the half sphere [tex]=2/3\pi r^3[/tex]

The volume of solid = volume of cone + volume of the half sphere

[tex]V=V_1+V_2[/tex]

Given

height h = 26 in

radius r = 8 in

[tex]V_1=1/3\pi r^2h[/tex]

    [tex]= 1/3 \times 3.14 \times 8 \times 8\times 26[/tex]

    [tex]=1741.65 \ in^3[/tex]

[tex]V_2=2/3\pi r^3[/tex]

 [tex]= 2/3 \times 3.14 \times 8 \times 8\times 8[/tex]

  [tex]= 1071.78 \ in^3[/tex]

[tex]V = 1741.65 \ in^3 + 1071.78 \ in^3[/tex]

[tex]2813 \ in^3[/tex]

Hence the volume of the solid is 2813 in³

Answer: 10

Step-by-step explanation:

500 divided 50

Therefore, the angle of depression from Lisa's line of sight to the cat is approximately 12.21 degrees.

An angle is a geometric figure formed by two rays (or line segments) that share a common endpoint, called the vertex. The rays are also called the sides or legs of the angle. Angles are usually measured in degrees, with a full circle measuring 360 degrees. Angles are used to describe the direction of an object, the orientation of a shape, or the position of a line. They are also used in trigonometry to calculate distances and heights in triangles, and in many other areas of mathematics and science.

Here,

In the diagram above, we have Lisa standing at point L and looking out the window of the building at her lost cat at point C, which is 58.5 meters away from the building. The height of the window from the ground is 12.6 meters. Let's call the angle of depression from Lisa's line of sight to the cat as ∠WLC. To find this angle, we need to use the tangent function, which relates the opposite side (12.6 meters) and adjacent side (58.5 meters) of ∠WLC.

tan ∠WLC = opposite / adjacent

tan ∠WLC = 12.6 / 58.5

Using a calculator, we get:

∠WLC = tan⁻¹(12.6 / 58.5)

∠WLC ≈ 12.21 degrees

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Answer:

$47.7

Step-by-step explanation:

You first have to find how much of 53 is 10%:  53/10 = 5.3  53-5.3= 47.7

Answer:

180 cubic centimeters

Step-by-step explanation:

We can find the volume of a box by multiplying the dimensions of the box together.
In other words

Volume = length × width × height

Given dimensions

By plugging these given dimensions into the formula we acquire

Volume = 6 × 10 × 3

==> multiply 10 and 6

Volume = 60 × 3

==> multiply 60 and 3

Volume = 180

The volume of the box is 180 cubic centimeters.

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The equation of the tangent line to the graph of y = xlnx at the point (2,2ln2) is y = (ln(2) + 1)x - 2ln2 - 2.

A tangent is a straight line or plane that touches a curve or curved surface at a single point, but does not intersect it at that point. The tangent line can be thought of as the instantaneous rate of change of the curve at that point. In calculus, the slope of the tangent line at a point on a curve is defined as the derivative of the curve at that point. The tangent line is often used to approximate the behavior of the curve near the point of tangency.

To find the equation of the tangent line to the graph of y = xlnx at the point (2,2ln2), we need to use the slope-point form of the equation of a line.

First, we find the slope of the tangent line by taking the derivative of the function y = xlnx:

y = xlnx

y' = ln(x) + 1

At x = 2, we have:

y' = ln(2) + 1

So the slope of the tangent line at (2,2ln2) is ln(2) + 1.

Next, we use the point-slope form of the equation of a line:

y - y1 = m(x - x1)

where (x1, y1) is the given point, and m is the slope we just found.

Substituting x1 = 2, y1 = 2ln2, and m = ln(2) + 1, we get:

y - 2ln2 = (ln(2) + 1)(x - 2)

Expanding and simplifying, we get the equation of the tangent line:

y = (ln(2) + 1)x - 2ln2 - 2

So the equation of the tangent line to the graph of y = xlnx at the point (2,2ln2) is y = (ln(2) + 1)x - 2ln2 - 2.

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Based on the dot plots, it can be concluded the students in the second period had higher scores with a typical score of 3 on the essay.

The dot plots show the scores Ms. Munoz students got on their essays for two different periods. In these plots, we can observe that in the first period the scores were between 1 and 4, while in the second period, the scores were between 2 and 5.

This shows the scores were higher in the second period, moreover, the typical or the most common score was 3.

Note: This question is incomplete; below I attach the missing information.

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Applying the angle relationships, the measures are: 1. m<A = 18°; m<B = 40°; 2. m<A = 9°; m<B = 43°; 3. m<C = 123°; m<D = 158°; 4. m<C = 131°; m<D = 160°

The position of the angle relationships in the image shown will determine the measures of the angles involved.

Therefore, we have the following for Figure  A:

1. m<D = 162°; m<C = 122°

m<A = 180 - 162 = 18° [the angle relationship is linear pair]

m<B = 180 - 122 - 18 = 40° [triangle sum theorem]

2. m<D = 171°; m<C = 128°

m<A = 180 - 171 = 9° [the angle relationship is linear pair]

m<B = 180 - 128 - 9 = 43° [triangle sum theorem]

3. m<A = 22°; m<B = 35°

m<C = 180 - 22 - 35 = 123° [triangle sum theorem]

m<D = 180 - 22 = 158° [the angle relationship is linear pair]

4. m<A = 20°; m<B = 29°

m<C = 180 - 20 - 29 = 131° [triangle sum theorem]

m<D = 180 - 20 = 160° [the angle relationship is linear pair]

Therefore, we have the following for Figure  B:

9. m<A = 65° [given]

m<B = 180 - 65 = 115° [linear pair]

m<C = 65° [corresponding angles]

m<D = m<B [corresponding angles]

m<D = 115°

m<E = m<C = 65° [vertical angles]

m<F = 180 - 65 - 90 = 25

10. m<A = 70° [given]

m<B = 180 - 70 = 110° [linear pair]

m<C = 70° [corresponding angles]

m<D = m<B [corresponding angles]

m<D = 110°

m<E = m<C = 70° [vertical angles]

m<F = 180 - 70 - 90 = 20

11. m<F = 22° [given]

m<E = 180 - 90 - 22 = 68°

m<E = m<C = 68° [vertical angles]

m<A = m<C = 68° [corresponding angles]

m<B = 180 - 68 = 112° [linear pair]

m<D = m<B = 112° [corresponding angles]

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The length of the missing leg of the triangle is 8 units. We have a right triangle ABC with angle BAC equal to 90 degrees. The length of one of the legs is 6 units, and the length of the hypotenuse is 10 units.

We are asked to find the length of the other leg of the triangle.

[tex]That is, a^2 + b^2 = c^2[/tex]

In this case, we know that the length of one leg is 6 units, and the length of the hypotenuse is 10 units. We can substitute these values into the Pythagorean theorem and solve for the missing leg:

[tex]a^2 + b^2 = c^2[/tex]

[tex]6^2 + b^2 = 10^2[/tex]

[tex]36 + b^2 = 100[/tex]

[tex]b^2 = 100 - 36[/tex]

[tex]b^2 = 64[/tex]

Taking the square root of both sides, we get:

b = 8

Therefore, the length of the missing leg of the triangle is 8 units.

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147

$3 x 147 cookies = (approximately) $440

$440 x 2 = 880

To the nearest hundredths place: 29.03 kg

Answer:

The probability of Meikel wearing khakis and sandals is

Step-by-step explanation:

3 ÷ 12 which should give us

1/4

The area of a parallelogram can be found by multiplying the base by the height

The formula for the area of a parallelogram is shown:

A = b × h

where A is the area, b is the base, and h is the height of the parallelogram.

Note that the base and the height must be perpendicular to each other.

1) A = 1/2bh

= 0.5 * 13/6 * 24

= 26 ft^2

1)A = bh

= 15/4 * 14/5

10.5 ft^2

3) 8/3 * x = 24

x = 24 * 3/8

x = 9 ft

4) x * 4/5 = 10

x = 10 * 5/4

= 12.5 ft

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The area of this triangle is equal to 26 yd².

The area of this parallelogram is equal to 21 ft² or 10 1/2 ft².

The base of this parallelogram as a fraction is 72/8 feet.

The base of this rectangle is 25/2 cm or 12 1/2 cm.

In Mathematics and Geometry, the area of a triangle can be calculated by using the following formula:

Area = 1/2 × base × height

Area = 1/2 × 24 × 2 1/6

Area = 12 × 13/6

Area = 156/6

Area = 26 yd²

In Mathematics and Geometry, the area of a parallelogram can be calculated by using the following formula:

Area of a parallelogram = base × height

Area of a parallelogram = 2 4/5 × 3 3/4

Area of a parallelogram = 14/5 × 15/4

Area of a parallelogram = 7 × 3/2

Area of a parallelogram = 21/2 = 10 1/2 ft²

24 = base × 2 2/3

24 = base × 8/3

Base = 72/8

Base = 9 feet.

For the base of the rectangle, we have:

Area of rectangle = length × breadth

10 = 4/5 × breadth

Breadth = 50/4

Breadth = 25/2 = 12 1/2 cm.

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The total cost for one admission and r rides is $10 for admission and $2 multiplied by r for the rides, which can be expressed as the formula: Cost = $10 + $2r.

The problem statement provides two pieces of information: the admission cost and the ride cost. We are asked to find the total cost for one admission and r rides.

The admission cost is straightforward - it is a fixed cost of $10. This means that regardless of how many rides you go on, you will always pay $10 just to get into the fair.

The ride cost is based on the number of rides you go on. Each ride costs $2, so if you go on r rides, you will pay $2 multiplied by r, or 2r, for the ride portion of the cost.

To find the total cost for one admission and r rides, we simply add the admission cost of $10 to the ride cost of 2r, which gives us the formula:

Cost = $10 + $2r

This formula tells us that the cost of one admission and r rides is equal to $10 plus $2 multiplied by the number of rides (r) that the person goes on.

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Answer:

-0.03334

Step-by-step explanation:

i just graphed it and looked at were y=4 and gave you the x value

:)

Answer:

x1 = -0.0333333, x2=0.7

Answer:

6.92 years

Step-by-step explanation:

To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where:

A = the final amount (in this case, $8800)

P = the initial investment (in this case, $4500)

r = the annual interest rate (in decimal form, so 6.8% = 0.068)

n = the number of times the interest is compounded per year (in this case, daily, so n = 365)

t = the number of years

We want to solve for t, so we can rearrange the formula as follows:

t = (ln(A/P)) / (n ln(1 + r/n))

where ln is the natural logarithm.

Plugging in the given values, we get:

t = (ln(8800/4500)) / (365 ln(1 + 0.068/365))

t ≈ 6.92

Therefore, it would take about 6.92 years (or about 7 years) for the investment to grow to $8800 at an annual interest rate of 6.8% compounded daily.

Answer:

A' (0, 0 )

Step-by-step explanation:

a translation of 5 units right adds 5 to the x- coordinate

a translation of 2 units down subtracts 2 from the y- coordinate

A (- 5, 2 ) → A' (- 5 + 5, 2 - 2 ) → A' (0, 0 )

Answer:

[tex] {≈225.4 \: m}^{2} [/tex]

[tex] {≈248.2 \: m}^{2} [/tex]

Step-by-step explanation:

(My first few steps are in the photo I've added)

Now, we can find the area of the first cutout:

[tex]s(cutout1) = \frac{\pi \times {r}^{2} \times \alpha }{360°} = \frac{\pi \times {18}^{2} \times 114°}{360°} = \frac{36963\pi}{360°} = 102.6\pi[/tex]

We can also find the altitude of the first triangle using trigonometry:

[tex] \sin(∠obd) = \frac{od}{ob} [/tex]

We also need to find AB by using the Pythagorean theorem:

[tex] {ad}^{2} = {ao}^{2} - {od}^{2} = {18}^{2} - {5.56}^{2} = 293.0864[/tex]

[tex]ad > 0[/tex]

[tex]ad = \sqrt{293.0864} ≈ 17.12[/tex]

[tex]ab = 2 \times 17.12 = 34.24[/tex]

[tex]h1 = od = ob \times \sin(∠obd) = 18 \times \sin(18°)≈5.56[/tex]

Then, we can find the area of the 1st triangle:

[tex]s (1triangle)= \frac{1}{2} \times 5.56 \times 34.24 ≈95.19[/tex]

And we can also find the area 1:

[tex]area1 = s(cutout1) - s(triangle1) = 102.6\pi - 95.19≈225.25[/tex]

Now let's do the same thing with the second area:

[tex]s(cutout2) = \frac{\pi \times {r}^{2} \times \alpha }{360°} = \frac{\pi \times {18}^{2} \times 131°}{360°} = \frac{42444\pi}{360°} =117.9 \pi[/tex]

[tex] \sin(∠obe) = \frac{oe}{oc} [/tex]

[tex]h2 = oe = oc \times \sin(∠obe) = 18 \times \sin(24.5°)≈7.46[/tex]

[tex] {be}^{2} = {bo}^{2} - {oe}^{2} = {18}^{2} - {7.46}^{2} = 268.3484[/tex]

[tex]be > 0[/tex]

[tex]be = \sqrt{268.3484} ≈16.38[/tex]

[tex]bc = 2 \times 16.38 = 32.76[/tex]

[tex]s(triangle2) = \frac{1}{2} \times 7.46 \times 32.76 = 122.19[/tex]

[tex]area2 = s(cutout2) - s(triangle2) = 117.9\pi - 122.19≈248.2[/tex]

I don't know if I got this right, though...

The area of the shaded segments in the circle is 422.5 square meters

Given that

Radius = 18 meters

Angles = 114 degrees and 131 degrees

The area of the shaded segments is the sum of the area of the individual segments and this is calculated as

Segment area =  (½) × r² × [(π/180) θ – sin θ]

Using the above, we have

Shaded segments = (½) × 18² × [(114π/180) – sin (114)] + (½) × 18² × [(131π/180) – sin (131)]

Evaluate

Shaded segments = 422.5 square meters

Hence, the shaded segments is 422.5 square meters

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Andre makes green paint by mixing 5 cups οf yellοw paint with 2 cups οf blue paint.

Tο make the same shade οf green, Andre needs 20 cups yellοw paint if he uses 8 cups οf blue paint.

Andre needs 6 cups blue paint if she uses 15 cups οf yellοw paint.

Any cοlοred liquid, liquefiable, οr sοlid mastic substance that fοrms intο a film when applied tο a surface in a thin layer is referred tο as paint. It can be used tο give texture, cοlοr, οr prοtectiοn, amοng οther things. There are numerοus types and shades οf paint that can be prοduced. Mοst paint kinds eventually sοlidify, but they are nοrmally kept, purchased, and applied as liquids. The bulk οf paints are either οil- οr water-based, and each has unique characteristics.

Andre makes green paint by mixing 5 cups οf yellοw paint with 2 cups οf blue paint.

Suppοse tο make the same shade οf green, he uses 8 cups οf blue paint and x cups οf yellοw paint.

Then, yellοw/blue= 5/2= x/8

Or, 2x= 5*8

Or, x= 40/2

Or, x=20

Sο tο make the same shade οf green, Andre needs 20 cups yellοw paint if he uses 8 cups οf blue paint.

Suppοse Andre needs y cups blue paint if she uses 15 cups οf yellοw paint.

Yellοw/blue= 5/2= 15/y

Or, 5y= 15*2

Or, y= 30/5

Or, y=6

Sο Andre needs 6 cups blue paint if she uses 15 cups οf yellοw paint.

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Answer:

Large box = 50 pounds

Small box = 15 pounds

Step-by-step explanation:

Because there are 50 small boxes, the truck is carrying 50 combined large and small boxes. That means the area of those boxes are 50 * 65 = 3250 pounds. There are 10 large boxes left. 3750-3250 = 500. The weight of 10 large boxes is 500 pounds, which means 1 large box is 500/10 which is 50 pounds.

If a large box is 50 pounds, and the combined weight is 65 pounds, 65-50 = 15 pounds for a small box.