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1 year ago

9

Y=-2/5x-3 in graph form

1 answer:

guapka [62]1 year ago

5 0

The points are (0, -3) and (5, -5)

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Factor 3x^2y^2 − 2xy^2 − 8y^2. Show your work.

docker41 [41]

<span>3x^2y^2 − 2xy^2 − 8y^2 =
y^2 (3x^2 - 2x - 8) =

factoring with leading coefficient:

for ax2+bx+c find two numbers n,m, that m*n = a*c and m+n = b
</span><span><span>
3x^2 - 2x - 8
a=3, b=-2, c=-8
</span>a*c = 3*(-8) = -24
-24=(-6)*4 and -6+4=-2, so m=-6 and n=4

replace bx with mx + nx and factor by grouping
</span><span>
3x^2 - 2x - 8 = </span>3x^2 -6x + 4x -8 = 3x(x-2) + 4(x-2) = (3x+4)(x-2)

answer:
<span>3x^2y^2 − 2xy^2 − 8y^2 = y^2(3x+4)(x-2)</span>

7 0

3 years ago

Find the P-value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject H0

Leviafan [203]

Answer: We do not reject the null hypothesis.

Step-by-step explanation:

When the p-value is greater than the significance level , then we do not reject the null hypothesis or if p-value is smaller than the significance level , then we reject the null hypothesis.

Given : Test statistic : $z = -2.28$

Significance level : $\alpha=0.02$

By using the standard normal distribution table ,

The p-value corresponds to the given test statistic ( two tailed ):-

$2P(Z>|z|)=2P(Z>2.28)=0.0226$

Since the p-value is greater than the significance level of 0.02.

Then , we do not reject the null hypothesis.

8 0

3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

2 years ago

REALLY confused easy question plz answer it will take 5 seconds

leonid [27]

Corresponding angles

3 0

2 years ago

If it’s too blurry let me know. I’ve been plugging in these points and idk maybe I’m doing something wrong but o can’t get the l

nika2105 [10]

Answer: A. -4

Step-by-step explanation:

6/((-4)^2-9)-(1/-4-3)= 6/7+1/7=1

4 0

3 years ago