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sertanlavr [38]
3 years ago
13

Can someone please help me I really need help please help me

Mathematics
1 answer:
just olya [345]3 years ago
7 0
Oh okay. I think that what this is asking, is that they just want you to find the slopes.

So first you have to put both equations into y=mx+b form

So first equation, x-2y+8=0, now we subtract 8 on both sides aswell as subtract x and were left with -2y=-x-8 now we have to divide by negative two on both sides which gives you y=1/2x+4.

Then we do the same to the second equation. 2x+y+1=0. Se switch some things around and get Y=-2x-1

Okay so your answer would be something like this.

These lines are perpendicular since the slopes are the inverse of one another. The second like would run in the direction of the 4 quadrant. And the first in the direction of the 1st quadrant.
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The wheels on Madison's bike are 64 inches in circumference. How many times do the wheels rotate if Madison rides 300 yards?
yuradex [85]

Answer:

Hello the answer is

diameter = 49cm;

radius = diameter/2 = 24.5cm;

circumference = 2πr ≈ 153.94 = 154cm

1 revolution = 154cm

50 revolution = 7700cm = 77m

I hope I am correct. also please mark brainliest

Step-by-step explanation:

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3 years ago
Factor the trigonometric equation.<br> cos2x-cosx
prisoha [69]

Cos(2x) = cos^2(x) - sin^2(x) - cos(x)

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cos(2x) - cos(x) = (2cos(x) + 1)(cos(x) - 1)

I think this is what you have asked for.

4 0
3 years ago
The length to width ratio of a wide-screen tv is 16:9. if a tv is 30 inches long, how wide is it?
bija089 [108]
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Karolina [17]

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Step-by-step explanation:

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Please help! I attached the question below.
kompoz [17]

Answer:

\frac{2(c+2)}{c(c-2)}

Step-by-step explanation:

\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }

Identity used:

a^{2}-b^{2}=(a-b)(a+b)

\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }

Now let us divide the modified expressions:

\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})} ÷ \frac{(c-2)^2}{2(6c^{3}+15c^{2}) }

we get:

\frac{2(c+2)}{c(c-2)}

5 0
3 years ago
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