Find a solution to the linear equation 9x+4y=−36 by filling in the boxes with a valid value of x and y.

Answer:

pay the 11 AM ticket

Step-by-step explanation:

Note that the flight last for 12 hours, and assuming the freelance developer can still work (have access to the internet) on the airplane throughout the flight, he stand to earn $960 ($80*12), which will still cover the cost of the flight with a profit of $60 ($960-900).

However, if he decides to pay the $11 flight ticket and there is no Internet on boards; there by losing a day of work, he stand to have lost working time which would earn with $900.

Therefore, the best choice is to pay the 11 AM ticket.

Answer:

(Going from left to right)

Box #1=3

Box #2=5

Box #3=7

Box #4=2

Step-by-step explanation:

For Box #4, there is nothing for the 2 to subtract from so it just goes down

For Box #3, it has to be 7, because nothing can be subtracted from 1 to get 3, so you would have to bring a 1 from the 4 to the left to make the 1 to a 10. 10-7=3

For Box #2 and 1, 3(we changed it in the last step) -9 = a negative number so we have to bring a 1 from the number to the left. This is a hard step but what you have to do it look at the bottom number, which is a 2, so that number had to be a 3 because 3-1=2. 4 becomes 14, and 14-9=5

Hope this helps, if it did, please consider giving me brainliest, it will help me a lot. If you have any questions, feel free to ask.

Have a good day! :)

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation of the answer:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

▬▬▬▬▬▬▬▬▬▬▬▬

Answer:

Angle 2= 45

Angle 3= 45

Angle 4= 135

Angle 5= 135

Angle 6= 45

Angle 7= 45

Angle 8= 135

Answer:

C ( 1,-2)

Step-by-step explanation:

We can plot the points and see what point is not in the shaded section

Answer:

The Man needs to run at 9 mph

Step-by-step explanation:

Let M stand for the man's speed in mph.  When the man  

runs toward point A, the relative speed of the train with respect  

to the man is the train's speed plus the man's speed (45 + M).  

When he runs toward point B, the relative speed of the train is the  

train's speed minus the man's speed (45 - M).

When he runs toward the train the distance he covers is 2 units.  

When he runs in the direction of the train the distance he covers  

is 3 units. We can now write that the ratio of the relative speed  

of the train when he is running toward point A to the relative speed  

of the train when he is running toward point B, is equal to the  

inverse ratio of the two distance units or

              (45 + M)          3

              -----------  =      ---

              (45 - M)          2

          90+2 M=135-3 M

⇒5 M = 45

⇒ M = 9 mph

The Man needs to run at 9 mph

Answer: 9 mph

Step-by-step explanation:

Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .

If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5

The remaining distance = 1 - 4/5 = 1/5

If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be

Speed = distance/time

Time = 1/5d ÷ speed

The time it will take the train to cover the entire distance d will be

Time = d ÷ 45

Equate the two time

1/5d ÷ speed = d ÷ 45

Speed = d/5 × 45/d

Speed = 9 mph

Answer:

h(g(x)) = x²+4x+4

Domain restriction = [tex][-\infty, \infty][/tex]

Step-by-step explanation:

Given the functions  h(x)=x^2 and g(x)=x+2, we are to find h(g(x)). To get the indicated operation we need to follow the steps;

Since the function in parenthesis g(x) = x+2

h(g(x)) can be written as h(x+2). Hence we are to look for the equivalent expression of  h(x+2).

Since h(x) = x², h(x+2) can simply be gotten by simply replacing the variable x in h(x) as x+2 as shown;

h(x+2) = (g(x))²

h(x+2) = (x+2)²

We can open the bracket

h(x+2) = x²+4x+4

The domain restriction is the point where the function cannot exist for the value of x. The function can therefore exist on any real value R. The only domain restriction is at the interval [tex][-\infty, \infty][/tex]

Hence h(g(x))  is equivalent to x²+4x+4.

Answer:

4x² + 30x - 2 = 0

Step-by-step explanation:

Given:

Area = 58 square feet

Width = 7 feet

Length = 8 feet

Since the area is 58, writing the equation, we have:

(8 + 2x)(7 + 2x) = 58

Now expand the equation:

56 + 16x + 14x + 4x² = 58

56 + 30x + 4x² = 58

Collect like terms:

30x + 4x² + 56 - 58 = 0

30x + 4x² - 2 = 0

Rearrange the equation to a proper quadratic equation:

4x² + 30x - 2 = 0

The quadratic equation that can be used to determine the thickness of the frame, x is 4x² + 30x - 2 = 0

Answer:

Area of the common pasture = 12 acres

Step-by-step explanation:

Let the dimensions of the farm owned by Jay are 'a' units and 'b' units.

Area of the farm = ab = 2 acres

Similarly, areas of the farm owned by Peter with dimensions 'a' unit and 'c' unit = ac = 4 acres

And area of the farm owned by Sam with dimensions 'b' and 'd' units = bd = 6 acres

Now, [tex]\frac{ab}{ac}=\frac{2}{4}[/tex]

[tex]\frac{b}{c}=\frac{1}{2}[/tex] ---------(1)

[tex]\frac{ab}{bd}=\frac{2}{6}[/tex]

[tex]\frac{a}{d}=\frac{1}{3}[/tex] ---------(2)

[tex]\frac{b}{c}\times \frac{a}{d}=\frac{1}{2}\times \frac{1}{3}[/tex]

[tex]\frac{ab}{cd}=\frac{1}{6}[/tex]

cd = 6(ab)

cd = 6 × 2 [Since ab = 2 acres]

    = 12 acres

Therefore, area of the common pasture will be 12 acres.

Answer:

800 artículos.

Step-by-step explanation:

Sea x = la cantidad de artículos vendidos

De la pregunta,

Precio de costo = 60 x + 4000

Venta (precio de venta) = 80x

También nos dieron ganancias como S12000

Para determinar la ganancia

Precio de venta: precio de costo = beneficio (cuando el precio de venta es más alto que el precio de costo)

Por lo tanto,

12,000 = 80x - (60x + 4000)

12,000 = 80x - 60x - 4000

Recolectando términos similares

12,000 + 4,000 = 80x - 60x

16,000 = 20x

x = 16,000 / 20

x = 800

Por lo tanto, la cantidad de artículos que se venderán si desea obtener una ganancia de S12,000 es de 800 artículos.

Answer:

x=18

Step-by-step explanation:

Answer:

x = 18

Step-by-step explanation:

Since 1/3x = 6,

x = 6 x 3

Thus, x = 18

Answer:

6 pads

Step-by-step explanation:

Take the amount of money and divide by the cost per pad

26/4 = 6 with 1 dollar left over

He can buy 6 pads

Answer:

$10 + $10 + $1 + 25¢ + 5¢

or

$20 + $1 + 25¢ + 5¢

Step-by-step explanation:

Each one must pay $21.30

Answer:

3/5

Step-by-step explanation:

5x= 3

x= 3/5

hope you understand the answer

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keep rocking

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Answer: 3rd Degree

Step-by-step explanation:

Answer:

Step-by-step explanation:

According to SOH in SOH, CAH TOA;

SOH means  sin theta = opposite/hypotenuse = 2/3

This shows that opposite = 2 and hypotenuse = 3. Before we can determine which of the expression is possible, we need to find the third side of the rigt angled triangle which is the adjacent.

According to Pythagoras theorem; hyp² = opp²+adj²

adj² = hyp² - opp²

adj² = 3² - 2²

adj² = 9 - 4

adj² = 5

adj = √5

Hence the adjacent side is √5.

From the trigonometry identity above;

cos theta = adj/hyp = √5/3 and tan theta = opp/adj = 2/√5

Since sec theta = 1/cos theta then sec theta = 1/(adj/hyp)

sec theta = hyp/adj = 3/√5

From the above calculation, the following are possible:

cos theta = √5/3, tan theta = 2/√5 and sec theta = 3/√5

The correct options are C and D

Answer: The first number is 4 and the second is 5

Step-by-step explanation:

Let's say the first number is x and the second is y:

x+2y=14

2x+y=13

I'll use the linear combination method to solve this.

  2x+4y=28

-  2x+y=13

3y=15

y = 5

x = 4

Hope it helps <3

Answer:

y=1

x=12

Step-by-step explanation:

Let the first number be x

Let the first number be y

Given:

x+2y=14

2x+y=13

Substitution:

x=14-2y

Solution:

2x+y=13

14-2y+y=13

-2y+y=13-14

-y=-1

y=1

x=14-2y

x=14-2(1)

x=14-2

x=12

Proof:

1) x+2y=14

12+2(1)=14

12+2=14

14=14

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer

Answer:

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.

Step-by-step explanation:

Vectorially speaking, let assume that ship A is located at the origin and the relative distance of ship B with regard to ship A at noon is:

[tex]\vec r_{B/A} = \vec r_{B} - \vec r_{A}[/tex]

Where [tex]\vec r_{A}[/tex] and [tex]\vec r_{B}[/tex] are the distances of ships A and B with respect to origin.

By supposing that both ships are travelling at constant speed. The equations of absolute position are described below:

[tex]\vec r_{A} = \left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]

[tex]\vec r_{B} = \left(170\,km\right)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]

Then,

[tex]\vec r_{B/A} = (170\,km)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j-\left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]

[tex]\vec r_{B/A} = \left[170\,km-\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]

The rate of change of the distance between the ship is constructed by deriving the previous expression:

[tex]\vec v_{B/A} = -\left(40\,\frac{km}{h} \right)\cdot i + \left(15\,\frac{km}{h} \right)\cdot j[/tex]

Its magnitude is determined by means of the Pythagorean Theorem:

[tex]\|\vec v_{B/A}\| = \sqrt{\left(-40\,\frac{km}{h} \right)^{2}+\left(15\,\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec r_{B/A}\| \approx 42.720\,\frac{km}{h}[/tex]

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.

Answer:

73

Step-by-step explanation:

Since DA is the diameter, this means that you can divide the circle into two half-circles.  You will solve with the right half-circle.  A half-circle has a 180 total degrees.  Add up the values given and solve for x.

(2x) + (7x - 4) + (85) = 180

2x + 7x - 4 + 85 = 180

9x - 4 + 85 = 180

9x + 81 = 180

9x = 99

x = 11

Now that you have the value of x, solve for m∠CZB.

m∠CZB = 7x - 4

m∠CZB = 7(11) - 4

m∠CZB = 77 - 4

m∠CZB = 73

Answer:

Pretty sure it is c

Step-by-step explanation:

Answer:

C.

Step-by-step explanation:

She will be painting the outsides of the table, so we need to find the surface area of the table.

There is the flat part of the table, which is a rectangular prism. There are also four legs, which are rectangular prism.

So, she will paint C. the surface area of 6 rectangular prisms.

Hope this helps!

Answer:

75%

Step-by-step explanation:

First we can solve the area of the rectangle originally the answer is:

4 × 2 = 8

Then we decrease both measurements by 50% to get the dimensions 1 and 2. The new area will be 1 × 2 which is 2.

2 is 25% of 8 which means that the area of the rectangle has been reduced by 75%.

Answer:

The variance of the data is 15.

σ² = 15

Step-by-step explanation:

The mean is given as

X = 8

х        |    (x - X)    |    (x - X) ²

12       |        4         |    16

9        |        1         |     1    

11        |        3         |    9

5       |        -3        |    9

3       |        -5        |    25

The variance is given by

[tex]\sigma^2 = \frac{1}{n-1} \sum (x - X)^2[/tex]

[tex]\sigma^2 = \frac{1}{5 - 1} (16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} ( 16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} (60) \\\\\sigma^2 = 15[/tex]

Therefore, the variance of the data is 15.

Answer: what is the question to this?

Step-by-step explanation: thanks let me know okay

Answer:

(11/5) (a ratio)

Step-by-step explanation:

11 ounces to 5 ounces could be rewritten as

 11 oz.

---------- = (11/5) (a ratio)

 5 oz

Answer:

the linear system has two valid solution.

Answer:one solution

Step-by-step explanation:

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

Answer:

(A) [tex]\text{The slope of OA is }\dfrac{4}{3}[/tex]

(C) It’s equation is 4x-3y=0.

Step-by-step explanation:

Point O is at (0,0)

Point A is at (3,4)

[tex]\text{Slope of OA}=\dfrac{4-0}{3-0} \\m=\dfrac{4}{3}[/tex]

The equation of a straight line is in the form: y=mx+b

The y-intercept of the line OA=0

Therefore, we have:

[tex]y=\dfrac{4}{3}x+0\\3y=4x+0\\$Subtract 3y from both sides$\\4x-3y=0[/tex]

The equation of the line is: 4x-3y=0

Answer:

X = [tex]\left[\begin{array}{ccc}2&-7&3\\13&0&2\end{array}\right][/tex]

Step-by-step explanation:

4B = -2X - 2A

Dividing both sides by -2

=> -2B = X + A

Subtracting A to both sides

=> X = -2B-A

Now, Let's Solve:

=> X = [tex]-2\left[\begin{array}{ccc}0&2&-2\\5&0&3\end{array}\right] -\left[\begin{array}{ccc}-2&3&1\\-3&0&4\end{array}\right][/tex]

=> X = [tex]\left[\begin{array}{ccc}-2*0&-2*2&-2*-2\\-2*5&-2*0&-2*3\end{array}\right] - \left[\begin{array}{ccc}-2&3&1\\-3&0&4\end{array}\right][/tex]

=> X = [tex]\left[\begin{array}{ccc}0&-4&4\\10&0&6\end{array}\right] - \left[\begin{array}{ccc}-2&3&1\\-3&0&4\end{array}\right][/tex]

=> X = [tex]\left[\begin{array}{ccc}0-(-2)&-4-3&4-1\\10-(-3)&0-0&6-4\end{array}\right][/tex]

=> X = [tex]\left[\begin{array}{ccc}2&-7&3\\13&0&2\end{array}\right][/tex]

Answer:

Product of given question is 6√5

Step-by-step explanation:

Given:

3√2(√10)

Find:

Product.

Computation:

⇒ 3√2(√10)

⇒ 3√20

⇒ 3√4×5

⇒ 3√2×2×5

⇒ 3×2√5

⇒ 6√5

Product of given question is 6√5

Answer:

The correct answer is:

Double-blind (b)

Step-by-step explanation:

A blind/blinded experiment is one in which information which may influence the participant or experimenter is withheld throughout the process of the experiment either by masking (giving false identity) or completely blinded, to avoid biases that may arise from such knowledge by the participant or experimenter.

Blinding is of three types: single-blind, double-blind and triple-blind experiements and this is named with respect to three categories involved in the experiment; participant, researcher or a third party, which may include: analysts, monitoring committees stakeholders etc. The blinding type is explained as follows

Blinding Type              participant       researcher           Third-party

single-blind                   blinded               unblinded           unblinded

double-blind                  blinded               blinded               unblinded

triple-blind                     blinded               blinded                blinded

In this example, the 200 headache sufferers (participants) and the Migrane testing service (researchers) do not know the contents of the bottles being administered, whereas the pharmaceutical company (third-party knows), hence it is a double-blinded experiment.

Answer:

C.I =  0.7608   ≤ p ≤   0.8392

Step-by-step explanation:

Given that:

Let consider a  random sample n = 400 candidates where  320 residents indicated that they voted for Obama

probability [tex]\hat p = \dfrac{320}{400}[/tex]

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to  develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

[tex]=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }[/tex]

where;

[tex]Z_{0.05/2}[/tex] = [tex]Z_{0.025}[/tex] = 1.960

SO;

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}[/tex]

[tex]=0.8 \pm 1.960 \times 0.02}[/tex]

[tex]=0.8 \pm 0.0392[/tex]

= 0.8 - 0.0392     OR   0.8 + 0.0392  

= 0.7608    OR    0.8392

Thus; C.I =  0.7608   ≤ p ≤   0.8392