Can anyone find the x and y intercepts for this equation?

1 answer:

7 0

1/3x - 1/2y = 1

At the 'x' intercept, y=0 , and the equation is 1/3 x = 1

Multiply each side by 3 : x = 3  <== the x-intercept

At the 'y' intercept, x=0, and the equation is -1/2 y = 1

Multiply each side by 2 : - y = 2

Multiply each side by -1 :  y = -2  <== the y-intercept

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I WILL REWARD BRAINLIEST!!!

kondor19780726 [428]

I believe it's the first 3 boxes

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3 years ago

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Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7

photoshop1234 [79]

Hope I can be of assistance!

$3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}$
$3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \ 3b^{-5}b^2c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ b^{-5}b^2=\:b^{2-5}=\:b^{-3}$
$3b^{-3}c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ c^{-4}c^7=\:c^{-4+7}=\:c^3$
$3b^{-3}c^3$

$\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \ b^{-3}=\frac{1}{b^3} \ \textgreater \ 3\cdot \frac{1}{b^3}c^3$

$\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:3c^3}{b^3}$

Finally
$\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{3c^3}{b^3}$

Hope this helps!

8 0

3 years ago

Please help! this is a final! FUNCTIONS

Gnom [1K]

The zeroes are -1/2 and -2
last option

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3 years ago

Factor. 21x^2 -32x+12

baherus [9]

Through trial and error:
(7x-6)(3x-2)

Usually it works best to test out a number of possibilities :)

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3 years ago