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almond37 [142]
3 years ago
13

Y= 3x+7 y = -2x-3 What would be the x and y for that solution

Mathematics
1 answer:
alexandr402 [8]3 years ago
6 0

Answer:

So if we are solving for x and y. First we need to subtract both equations. We take each equation and put in parenthesis and subtract.

y-y = (3x+7)-(-2x-3) =

0 = 3x+7+2x+3

0 = 5x+10

-10 = 5x

-2 = x.

Now we need to solve for y.

Plug it in for the first equation. -2*3+7 = -6+7 = 1

Test if we are correct and try in the second equation.

-2*-2-3 = 4-3 = 1

<u>x = -2</u>

<u>y = 1</u>

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A triangle has one leg that measures 12 cm and a hypotenuse 24 cm. Find the length of the other leg to the nearest tenth of cent
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Answer:

  20.8 cm

Step-by-step explanation:

The term "hypotenuse" suggests this triangle is a right triangle. Then the other leg can be found using the Pythagorean theorem:

  x^2 +12^2 = 24^2

  x^2 = 576 -144 = 432

  x = √432 ≈ 20.8 . . . cm

The other leg is about 20.8 cm.

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<em>Additional comment</em>

A right triangle with a short leg that is 1/2 the length of the hypotenuse is the "special" 30°-60°-90° right triangle. The longer leg is √3 times the length of the short leg: 12√3 ≈ 20.8 cm.

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Find the zeros, vertical asymptote, horizontal asymptote, slant asymptote, and graphing points.
kramer
Zeroes
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)

HA

for p(x)/q(x)

when degree of p(x)<q(x), HA=0

when degree of p(x)=q(x), HA= leading coef of p(x) divided by leading coef of q(x)

when degree of p(x)>q(x) you probably have a slant assymtote

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leading coefs

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does it cross?

1=(x^2)/(x^2-4)

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minus x^2 both sides

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VA

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set the denomenator to zero

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VA's at x=2 and x=-2




so to graph

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draw the lines

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x=-2

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so from left, it goes from above the HA right up to the VA of x=-2

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