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3 years ago

Find the product. -2x(x2 - 3) -2 x3 - 6 x -2 x2 6 x 6 x3 -2 x3 6 x.

1 answer:

6 0

Product of two polynomials can be done by distribution of multiplication over addition. The product of $-2x(x^2 - 3)$ is $-2x^3 + 6x$

<h3>What is distributive property of multiplication over addition?</h3>

Suppose a, b and c are three numbers. Then we have:

$a(b + c) = a\times b + a\times c$

(a(b+c) means a multiplied to (b+c). The sign of multiplication is usually hidden when using symbols and both quantities which are in multiplication are written together without space)

The product result of the given expression is

$-2x(x^2 - 3) = -2x^{1 + 2} + 6x = -2x^3 + 6x$

It is because when bases are same and there is multiplication, then powers add up, so we have: $a^b \times a^c = a^{b+c}$

Thus,

The product of $-2x(x^2 - 3)$ is $-2x^3 + 6x$

Learn more about product of polynomials here:

brainly.com/question/9106484

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Help me with this I’ll give you 5 stars and brainlyest

Elan Coil [88]

Answer:

Step-by-step explanation:

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5 0

4 years ago

A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to

Paladinen [302]

Answer:

Step-by-step explanation:

Hello!

The objective of the research is to compare the newly designed drug to reduce blood pressure with the standard drug to test if the new one is more effective.

Two randomly selected groups of subjects where determined, one took the standard drug (1- Control) and the second one took the new drug (2-New)

1. Control

X₁: Reduction of the blood pressure of a subject that took the standard drug.

n₁= 23

X[bar]= 18.52

S= 7.15

2. New

X₂: Reduction of the blood pressure of a subject that took the newly designed drug.

n₂= 21

X[bar]₂= 23.48

S₂= 8.01

The parameter of study is the difference between the two population means (no order is specified, I'll use New-Standard) μ₂ - μ₁

Assuming both variables have a normal distribution, there are two options to estimate the difference between the two means using a 95% CI.

1) The population variances are unknown and equal:

[(X[bar]₂-X[bar]₁)± $t_{n_1+n_2-2;1-\alpha /2}$ *( $Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ )]

$t_{n_1+n_2-2;1-\alpha /2}= t_{23+21-2;1-0.025}= t_{42;0.975}= 2.018$

$Sa=\sqrt{\frac{(n_1-1)*S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{22*7.15^2+20*8.01^2}{42} }= 7.57$

[23.48-18.52]±2.018*( $7.57*\sqrt{\frac{1}{21} +\frac{1}{23} }$ )]

[0.349; 9.571]

2) The population variables are unknown and different:

Welche's approximation:

[(X[bar]₂-X[bar]₁)± $t_{Dfw;1-\alpha /2}$ *( $\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} }$ )]

$Df_{w}= \frac{(\frac{S_1^2}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1}+ \frac{(\frac{S_2^2}{n_2} )^2}{n_2-1} } = \frac{(\frac{7.15^2}{23} +\frac{8.01^2_2}{21} )^2}{\frac{(\frac{7.15^2}{21} )^2}{20}+ \frac{(\frac{8.01^2}{23} )^2}{22} } = 42.85= 42$

$t_{Df_w;1-\alpha /2}= t_{42; 0.975}= 2.018$

[(23.48-18.52)±2.018 $\sqrt{\frac{7.15^2}{23} +\frac{8.01^2}{21} }$ ]

[0.324; 9.596]

I hope this helps!

4 0

3 years ago

Given ZW=20. Answer the questions please.

djyliett [7]

a) Corresponding angles of similar figures are congruent, so $\angle B \cong \boxed{\angle Z}$

b) Scale factor = (image)/(preimage). Since we are looking for ABCD to WZYX, this means that WZYX is the image and ABCD is the preimage. Thus, the scale factor is $\frac{25}{10}=\boxed{5/2}$