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lina2011 [118]
2 years ago
12

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
7 0
3 years ago
The graph shows the relationship between a gift card balance and the number
faltersainse [42]

Answer:

3

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.

5 0
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
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