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2 years ago

If y = 12 when x = –15, find y when x = 2.

Mathematics

1 answer:

Shtirlitz [24]2 years ago

4 0

Answer:

Step-by-step explanation:

First make an equation that shows the relationship between x and y by finding what you need to do to x to get y:

-15 + ? = 12

? = 27

so the equation is : y = x + 27

now just plug in 2 for x:

y = 2 + 27

y + 29

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Is negative three a perfect cube?

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2 years ago

Jennifer was baking cupcakes for her class. She needed to bake 35 cupcakes in all. Each cupcake needed .13 cups of sugar. How ma

Paraphin [41]

Answer and working out attached below. Hope it helps

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2 years ago

drag each expression to the correct location on the graph and equation. not all expressions will be used. complete the given dia

Svetllana [295]

What we have here in the picture is a graph of a segment. The goal is to find the segment's length, which we can do by using the distance formula:

$d^2=(x_2-x_1)^2-(y_2-y_1)^2$

Note that the distance between two points is equal to the length of the line segment that joins them.

So now we already have answers for two of the boxes.

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So, the other two boxes should contain:

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9 months ago

If cot theta= 4/3, find csc theta

Marrrta [24]

Answer: Option d.

Step-by-step explanation:

The trigonometric identity needed is:

$csc^2\theta=cot^2\theta+1$

Knowing that $cot\theta=\frac{4}{3}$ :

Substitute it into $csc^2\theta=cot^2\theta+1$ :

$csc^2\theta=(\frac{4}{3})^2+1$

Simplify the expression:

$csc^2\theta=(\frac{4}{3})^2+1\\\\csc^2\theta=\frac{16}{9}+1\\\\csc^2\theta=\frac{25}{9}$

Solve for $csc\theta$ . Apply square root at both sides of the expression:

$\sqrt{csc^2\theta}=\±\sqrt{\frac{25}{9}}$

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8 0

2 years ago

An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15

NeTakaya

Answer:

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

$sin 42^0 = \frac{Opposite}{Hypotenuse} \\\\Sin42^0 = \frac{15000}{x}\\ \\x = \frac{15000}{sin42^0}\\\\ \\$

$x = \frac{15000}{ 0.6691} \\\\x = 22,417 feet$

Hence the distance (x) from the plane P to the observer O is approximately 22,417 feet

3 0

3 years ago