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2 years ago

What are the x- and y-intercepts of −6x+4y=96 ?

1 answer:

3 0

* intercepts "x" --> y=0

-6x + 4(0) = 96
-6x = 96
x = $- \frac{96}{6}$

x = 16

(16,0)

*intercepts "y" --> x=0

-6(0) + 4y = 96
4y = 96
y = $\frac{96}{4}$

y = 24

(0,24)

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For what value of k are the roots of the quadratic equation kx (x-2)+6=0 equal?

ser-zykov [4K]

Answer:

$\boxed{\sf k=6}$

Step-by-step explanation:

$\sf kx (x-2)+6=0$

Expand brackets.

$\sf kx^2 -2kx+6=0$

This is in quadratic form.

$\sf ax^2 +bx+c=0$

Since this is for equal roots:

$\sf b^2 -4ac=0$

$\sf a=k\\b=-2k\\c=6$

$\sf (-2k)^2 -4(k)(6)=0$

$\sf 4k^2-24k=0$

$\sf 4k(k-6)=0$

$\sf 4k=0\\k=0$

$\sf k-6=0\\k=6$

Plug k as 0 to check.

$\sf \sf 0x^2 -2(0)x+6=0\\6=0$

False.

So that means k must equal 6.

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3 years ago

Planes A and B are shown.

Gnom [1K]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

Sin(60-theta)sin(60+theta)

ehidna [41]

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$