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3 years ago

What are the solutions to the equation (x-2)(x + 5) = 0?

Mathematics

1 answer:

Aneli [31]3 years ago

3 0

Answer:

Step-by-step explanation:

A product of two (or more) factor can be zero if and only if at least one of the factors is zero.

In other words, you cannot multiply two non-zero real numbers, and have zero as a result.

So, if we want the product of these two factors to be zero, at least one of them has to be zero.

The first factor is zero if

x-2=0 x=2

The second factor is zero if

x + 5 = 0 x= -5

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Which College entrance exam has the least amount of math problems to work?

taurus [48]

Answer:

SAT

Step-by-step explanation:

The two main college entrance exams are the SAT and ACT which are used for almost all colleges in the United States. Between these two the SAT has a lesser focus on math in general. Instead, the SAT tends to focus more on word problems and the process taken to solve them. This does not mean that it has no math at all, there is still a math section on the SAT's but it is not as extensive as in the ACT's.

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3 years ago

Complete the table below to give the outcomes and the corresponding probabilities<br>

matrenka [14]

Put a photo or something so we can answer

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2 years ago

I have A and B, I need help with C.

WARRIOR [948]

The answer is $382.80. the perimeter is 600ft. if fence post are placed every five feet, you need 120 post because 600÷5 is 120. each post cost 3.19 so you need 382.80 because 3.19×120 is 382.8

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3 years ago

A cylindrical plastic pipe’s length is 12 cm and it’s radius is 2.5 cm. To the nearest tenth, what is the volume of the pipe?

NemiM [27]

Answer:

V = 235.6 cm³

Step-by-step explanation:

the formula for the volume of a cylinder of radius r and length h is

V = πr²h.

Here, r = 2.5 cm and h = 12 cm, so the volume is:

V = π(2.5 cm)²(12 cm) = 75π cm³

To the nearest tenth, this volume would be V = 235.6 cm³

7 0

3 years ago

Read 2 more answers

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