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3 years ago

G(x) = f(x - 4) + 5 Transformation(s):

1 answer:

6 0

To find the transformation, compare the function to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.
Vertical compression

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6x^2-5x-4=0 <br> solving quadratics by factoring (answer with {} around it)

stepan [7]

Step-by-step explanation:

6x²–5x–4 = 0

6×4 = 24..

write many ways 24 can be written..

24 = (6×4), (1×24), (12×2), (8×3)

(8×3) --> –8+3 = –5

6x²+3x–8x–4 = 0

6x(x+½)–8(x+½) = 0

(6x–8)(x+½) = 0

6x–8 = 0, x+½ = 0

6x = 8, x = –½

x = 4/3, –½

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3 years ago

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Serga [27]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportat

Mrrafil [7]

Answer:

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 120$

Alternative hypothesis: $\mu \neq 120$

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z$

Step-by-step explanation:

Data given and notation

$\bar X=114$ represent the sample mean

$\sigma=22$ represent the population standard deviation

$n=36$ sample size

$\mu_o =120$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part 1

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 120$

Alternative hypothesis: $\mu \neq 120$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z$

8 0

3 years ago

What is the value of the expression 24 + 3²? What is the value of the expression 24 + 3² ?

timurjin [86]

Answer:

Step-by-step explanation:

3² = 3.3 = 9 => 24 + 9 = 33

3 0

2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago