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2 years ago

Write y = 2/5 x+ 5 in standard form

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

7 0

Answer:

-2x + 5y = 25

Step-by-step explanation:

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Tristan invested $470

Vlad1618 [11]

Well Tristan better get it together

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3 years ago

The graph shows the relationship between a gift card balance and the number

faltersainse [42]

Answer:

Step-by-step explanation:

We can see that the graph is constant.=> the path of the dot/ the pattern is constant or not changing.

The balance was 18 when there were no visits.

It was 0 after 6 visits.

=>the amount spent on each visit is

Balance/number of visits= 18/6= $3

This can clearly be seen on the graph.

It may not work with a graph where the patter is not constant/ keeps on changing.

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2 years ago

Read 2 more answers

A. Ecuación ordinaria, radio y ecuación general de la circunferencia con centro en (2,0) y tangente a la recta 4x+3y+7=0

Alexandra [31]

A is the answer my friend

7 0

2 years ago

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

2 years ago

Can someone help me with this I don’t understand

Komok [63]

Answer:

attached

Step-by-step explanation:

You should be able to do the 2nd problem yourself.

5 0

3 years ago