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3 years ago

Rezolvati 1) -2x+1= -7 2) 6+3x= -54+x 3) 2x+9= 7 +x 4) 3-2x= -x -6 5) -3(x+1) = -6 6) -4 (x -2)=12 7) -7(x-3)= -14

Mathematics

1 answer:

Natalija [7]3 years ago

3 0

<span>Răspunsurile sunt

1. x= 4
2. x = -30
3. x = -2
4. x = 9
5. x = 1
6. x = -1
7. x = 5
</span>

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Factor:

Naddik [55]

Answer:

D) (5x - 3) (25x^2+ 15x + 9)

Step-by-step explanation:

$125 {x}^{3} - 27 \\ \\ = ( {5x)}^{3} - {(3)}^{3} \\ \\ = (5x - 3) \{(5x) ^{2} + 5x \times 3 + ( {3})^{2} \}\\ \\ = (5x - 3) \{25 {x}^{2} + 15x + 9\}$

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3 years ago

For the function ƒ(x) = x2 – 4, find the value of ƒ(x) when x = 6.

pashok25 [27]

Answer:

f(x)= 2x -4

when x=6

f(x) = 2(6)-4

=12-4

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3 years ago

Read 2 more answers

Write a equation of a hyperbola given the foci and the asymptotes

professor190 [17]

Solution:

The standard equation of a hyperbola is expressed as

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\text{ \lparen parallel to the x-axis\rparen} \\ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ \lparen parallel to the y-axis\rparen} \end{gathered}$

Given that the hyperbola has its foci at (0,-15) and (0, 15), this implies that the hyperbola is parallel to the y-axis.

Thus, the equation will be expressed in the form:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ ----equation 1}$

The asymptote of n hyperbola is expressed as

$y=\pm\frac{a}{b}(x-h)+k$

Given that the asymptotes are

$y=\frac{3}{4}x\text{ and y=-}\frac{3}{4}x$

This implies that

$a=3,\text{ and b=4}$

To evaluate the value of h and k,

$undefined$

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1 year ago

Use the drawing tool(s) to form the correct answers on the provided grid. Consider the function g. For the x-value given in the

NeX [460]

Answer:

(-2,-12), (-1,-6), (0,-3) and (1,-3/2)

Step-by-step explanation:

g(x) = -3(1/2)^x

Putting values of x

x g(x)

-2 -3(1/2)^-2 = -12

-1 -3(1/2)^-1 = -6

0 -3(1/2)^0 = -3

1 -3(1/2)^1 = -3/2

Now, making the graph we will plot

(-2,-12), (-1,-6), (0,-3) and (1,-3/2)

The graph is shown in figure below.

8 0

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