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3 years ago

Choose the correct simplification of (6x-5)(2x2-3x-6)

Mathematics

2 answers:

seraphim [82]3 years ago

6 0

Answer:

The correct simplification of $(6x-5)(2x^2-3x-6)$ is:

$12x^3-28x^2-21x+30$

Step-by-step explanation:

we have to evaluate:

$(6x-5)(2x^2-3x-6)\\\\=6x(2x^2-3x-6)-5(2x^2-3x-6)\\\\=12x^3-18x^2-36x-10x^2+15x+30\\\\=12x^3-28x^2-21x+30$

Hence, correct simplification of $(6x-5)(2x^2-3x-6)$ is:

$12x^3-28x^2-21x+30$

Lady bird [3.3K]3 years ago

5 0

I think it equals 12x3−28x2−21x+30
hope this helps!

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adell [148]

First what you wanna do is get y by it’s self. To do that is to first add 5 on each side.

-5+3y=-4
+5 +5

3y= 1

Than you divide 3 on each side and it would give you y=1/3 if it’s a fractions problem but if you looking for decimals you would have to divide 1 by 3 which gets you .33333333.

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3 years ago

Which is an equation of the line that has a slope of and passes through (-1,3)?

zavuch27 [327]

Answer:

the second one is the right answer (2)

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3 years ago

"A cylindrical container has a radius of 25 inches and a height of 31 inches. What is the volume of

photoshop1234 [79]

Use the formula:

Radio= r, Height= h

V= 3.14xrx2h

r =25, h=31

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V= 4867 cubic inches

Step-by-step explanation:

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3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Which expression is equivalent??? help!

dexar [7]

Answer:

The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

Step-by-step explanation:

Given:

$\sqrt[3]{256x^{10}y^{7} }$

Solution:

We will see first what is Cube rooting.

$\sqrt[3]{x^{3}} = x$

Law of Indices

$(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}$

Now, applying above property we get

$\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})$

∴ The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

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3 years ago