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2 years ago

Helppp PLEASEEEE!!!!!! !!!

Mathematics

1 answer:

Ronch [10]2 years ago

6 0

Answer:

-3hf

Step-by-step explanation:

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MULTIPLE CHOICE ASAP !!!!!!!

grin007 [14]

The coefficient of x^2 is positiv so the parabola opens upwards.
whritten in vertex form its ( x + 2)^2
so the line of symmetry is x = -2 and the vertex is at (-2,0)
and as uts y>0 the curve will be dashed.
and the shading will be inside

so all are true except the last one

4 0

3 years ago

Read 2 more answers

I need to write a paragraph on setting telling how tree he author develops and use the setting I need to start with

Mars2501 [29]

Setting the scene for a story means giving your reader interesting details of time and ... Take, for example, David Mitchell's scene-setting near the start of his novel i want you to be its conscience, the eyes in the trees.

5 0

2 years ago

AB is the bisector of CAD. If the mCAB = 2x+30 and m DAB=5x+12 what is the mCAB

Dima020 [189]

Answer:

$CAB = 42$

Step-by-step explanation:

Given

$CAB = 2x + 30$

$DAB = 5x + 12$

Bisector: AB

Required

Find CAB

Since AB bisects CAD, then

$CAB = DAB$

This is so because AB divides CAD into two equal parts which are CAB and DAB

$CAB = DAB$

Substitute values for CAB and DAB

$2x + 30 = 5x + 12$

Collect Like Terms

$2x- 5x = 12 - 30$

$-3x = -18$

Divide both sides by -3

$x = 6$

To solve for CAB, we simply substitute 6 for x in $CAB = 2x + 30$

$CAB = 2 * 6 + 30$

$CAB = 12 + 30$

$CAB = 42$

3 0

2 years ago

Use the bionomial theorem to write the binomial expansion

MaRussiya [10]

Answer:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

Step-by-step explanation:

$\left(\frac{1}{2}x+3y \right)^4=\left(\frac{x}{2}+3y \right)^4\\$

Binomial Expansion Formula:

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ , also $\binom{n}{k}=\frac{n!}{(n-k)!k!}$

We have to solve $\left(\frac{x}{2} + 3 y\right)^{4}=\sum_{k=0}^{4} \binom{4}{k} \left(3 y\right)^{4-k} \left(\frac{x}{2}\right)^k$

Now we should calculate for $k=0, k=1, k=2, k=3 \text{ and } k =4;$

First, for $k=0$

$\binom{4}{0} \left(3 y\right)^{4-0} \left(\frac{x}{2}\right)^{0}=\frac{4!}{(4-0)! 0!}\left(3 y\right)^{4} \left(\frac{x}{2}\right)^{0}=\frac{4!}{4!}(81y^4)\cdot 1 =81 y^{4}$

It is the same procedure for the other:

For $k=1$

$\binom{4}{1} \left(3 y\right)^{4-1} \left(\frac{x}{2}\right)^{1}=54 x y^{3}$

For $k=2$

$\binom{4}{2} \left(3 y\right)^{4-2} \left(\frac{x}{2}\right)^{2}=\frac{27}{2} x^{2} y^{2}$

For $k=3$

$\binom{4}{3} \left(3 y\right)^{4-3} \left(\frac{x}{2}\right)^{3}=\frac{3}{2} x^{3} y$

For $k=4$

$\binom{4}{4} \left(3 y\right)^{4-4} \left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{16}$

You can perform the calculations, I will not type everything.

The answer is the sum of elements calculated.

Just organizing:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

8 0

2 years ago

Read 2 more answers

The following hypotheses are given. H0 : σ1² ≤ σ2² H1 : σ1² > σ2² A random sample of five observations from the first populat

Harrizon [31]

Answer:

Desision is to not reject the null hypothesis, this means that the variance of the second population is equal or greater than the variance of the first population.

Step-by-step explanation:

Hello!

You have the hypothesis set:

H₀: δ₁² ≤ δ₂²

H₁: δ₁² > δ₂²

α: 0.01

To study the population variance of two populations you have to work with the F-distribution. The statistic is:

F= (S₁²/S₂²)*(σ₁²/σ₂²) ~ F $_{(n1-1);(n2-1)}$

This is a one tailed test, the critical value is

$F_{(n1-1); (n2-1); 1-\alpha } = F_{4; 6; 0.99} = 9.15$

If the statistic value F $_{H0}$ ≥ 9.15, the decision is to reject the null hypothesis.

If the value F $_{H0}$ < 9.15, the decision is to not reject the null hypothesis.

F $_{H0}$ = (S₁²/S₂²)*(σ₁²/σ₂²) = (12/7) * 1

F $_{H0}$ = 1.714

Desision is to not reject the null hypothesis, this means that the variance of the second population is equal or greater than the variance of the first population.

I hope you have a SUPER day!

8 0

2 years ago