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mrs_skeptik [129]
3 years ago
5

Y2 – 8y + 16 = 64 how would i factor this equation?

Mathematics
1 answer:
Talja [164]3 years ago
3 0

Answer: y= 4 or y= -12

Step-by-step explanation:

Y²-8y+16=64.

Rearrange

Y²-8y+16-64=0

Y²-8y-48=0

(Y²+12y)-(4y-48)=0

Y(y+12)-4(y+12)=0

(Y-4)=0 or (y+12)=0

Y-4=0 or y+12=0

Y=4 or y=-12

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Thepotemich [5.8K]

Answer:

I notice that it looks like a bunch of tiny triangles. I wonder why there is only one chip with hot on it. It reminds me of a pie chart used to write down data.

Step-by-step explanation:

hope that helps you :)

5 0
3 years ago
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cookies are sold singly or packages of 6 or 18. with this packaging, how many ways can you buy 36 cookies?. Help Please!
ki77a [65]
<span>Ok can't count properly, lets try again one package has 18,
how many ways?

18, 18
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Therefore, there are 5 ways to buy 36 cookies.

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6 0
3 years ago
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4) -4y= -x + 4<br> 6) y + 3x = 3
wel

Answer:

x=16/13

y= -9/13

Step-by-step explanation:

A:         -4y=-x+4

B:          y+3x=3

4B:       4y+12x=12

A+4B:   12x= -x+16

----->    13x=16

               x=16/13

-4y= -16/13+4

4y=16/13-4

y=4/13-1= -9/13

4 0
3 years ago
Find the slope and y-intercept for 2x+8y=-32 and then graph the line
sammy [17]
2x+8y=-32
8y=-2x-32
y=-1/4x-4

The slope is -1/4, the y-intercept is -4, and I can't graph it for you on here. Type my equation into a graphing calculator and it'll do it for you. (Desmos is a good online graphing calc, and it's free)
8 0
4 years ago
Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you
Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi, we should expect \cos x. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi

so we should expect \tan\dfrac x2 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

3 0
3 years ago
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