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3 years ago

14

1.) If x varies directly as y and x = 35 when y = 7. What is the value of y when x = 25?

2 answers:

Fiesta28 [93]3 years ago

8 0

Answer:

y = 5

Step-by-step explanation:

$x \ \alpha \ y$

$=> x = k \times y\\=> 35 = k \times 7\\=> k = 5$

$x = k \times y$

$=>25 = 5 \times y\\ =>5 = y$

Temka [501]3 years ago

7 0

Answer:

y = 5

Step-by-step explanation:

x = 35, y = 7

x = ?y

35 = ?7

? = 5

x = 25, y = ?

x = ?y

25 = 5y

y = 5

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what is the rate of change for the linear relationship modeled in the table? x y −1 10 1 9 3 8 5 7 -1/2 0 1/2 2

Vesnalui [34]

Answer:

Rate of change for the linear relationship modeled is $\dfrac{-1}{2}$

Step-by-step explanation:

As the there is a linear relationship in the points, so all these points will be on a single straight line. Hence the slope will be same throughout all the points.

We know that, the slope of the line joining (x₁, y₁) and (x₂, y₂) is,

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Putting the points as (-1, 10) and (1, 9), we get

$m=\dfrac{9-10}{1-(-1)}$

$=\dfrac{9-10}{1+1}$

$=\dfrac{-1}{2}$

Rate of change is the slope of the line joining all these points.

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3 years ago

Given the function {f(x)=32-x}^{2}f(x)=32−x 2 , what is f(-5)f(−5)?

erastova [34]

Step-by-step explanation:

the description is very difficult to understand with special characters and potential typing errors.

I suspect the definition is

function(x) = 32 - x²

function(-5)function(-5) is then very easily calculated.

we need to calculate function(-5) and then square the result, as function(-5)function(-5) is just a simple multiplication of the function results.

function(-5) = 32 - (-5)² = 32 - 25 = 7

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Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

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3 years ago