If f(x)=5x-25 and g(x)=1/5x+5, which expression could be used to verify g(x) is the inverse of f(x)?

<img src="https://tex.z-dn.net/?f=211%3D6%283%2B8x%29%2B1" id="TexFormula1" title="211=6(3+8x)+1" alt="211=6(3+8x)+1" align="abs

harkovskaia [24]

X=4
211=18+48x+1
211=19+48x

5 0

3 years ago

Read 2 more answers

Diego’s current age is five times Martina’s age ten years ago. If Martina is currently m years old, what is Diego’s current age

puteri [66]

Answer:

5m-50

Step-by-step explanation:

Let Diego current age be x,

Given that Diego’s current age is five times Martina’s age ten years ago

Martina current age = m years

10 years ago Martina's age = m-10

Hence we have

x = 5(m-10) = 5m-50

So we find that Diego's current age can be expressed in terms of m as

Deigo's current age= 5m-50

Verify:

Deigo current age = 5m-50

Hence Martina age 10 years ago = 5m-50 divided by 5 = m-10

Martina current age = m-10+10 = m

Verified.

8 0

3 years ago

AB is dilated by a scale factor of 3 to form AB . Point O is the center of dilation, and point O lies on AB . If the slope of AB

avanturin [10]

<h2>Answer:</h2>

Option: D is the correct answer.

D. The slope of is 3, and passes through O.

<h2>Step-by-step explanation:</h2>

It is given that:

AB is dilated by a scale factor of 3 to form AB .

Also, the line AB pass through the origin.

So, if AB pass through (0,0) and some point (x,y) then the slope of line AB is:

$Slope=\dfrac{y-0}{x-0}\\\\i.e.\\\\Slope=\dfrac{y}{x}=3$

( Since, it is given the slope of AB is 3 )

Now , if AB is dilated to A'B' by a scale factor of 3 then

(0,0) → (0,0)

and

(x,y) → (3x,3y)

i.e. (0,0) and (3x,3y) will lie on A'B'.

Hence, the slope of line A'B' is given by:

$Slope=\dfrac{3y-0}{3x-0}\\\\i.e.\\\\Slope=\dfrac{3y}{3x}\\\\i.e.\\\\Slope=\dfrac{y}{x}=3$

Hence, both have the same slope.

The answer is : Option: D

5 0

4 years ago

I’ll give you brainlest if you answer this question correctly!

slamgirl [31]

Answer:

(5,4)

Step-by-step explanation:

multiply each equation by the value that makes the coefficients of x opposite

(-1) · (4x + 6y) = (-1) · 44

4x + 2y = 28

simplify

-4x - 6y = -44

4x + 2y = 28

add the two equations together to eliminate x from the system

-4x - 6y = -44

-4y = -16

divide each term by -4

y = 4

substitute the value for y into one of the original equations, then solve for x

-4x - 6y = -44

-4x - 6(4) = -44

-4x - 24 = -44

+ 24 + 24

-4x = -20

divide by -4

x = 5

so, since x = 5 and y = 4 ....

the solution to the independent system of equations can be represented as a point.

(5,4)

hope this helps! :)

5 0

3 years ago

Researchers obtained a sample of n = 10 children who attended daycare before starting school. The children are given a standardi

attashe74 [19]

Answer:

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

$p_v =2*P(t_{9}>4.093)=0.00084$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

Step-by-step explanation:

Data given and notation

53, 57, 61, 49, 52, 56, 58, 62, 51, 56

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X))^2}{n-1}}$

And we got:

$\bar X=55.5$ represent the sample mean

$s=4.249$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =50/tex] represent the value that we want to test [tex]\alpha=0.05$ represent the significance level for the hypothesis test (ASSUMED).

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal or not to 50 :

Null hypothesis: $\mu = 50$

Alternative hypothesis: $\mu \neq 50$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

Calculate the statistic

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

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

P-value

We need to calculate the degrees of freedom first given by:

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

Since is a two-sided tailed test the p value would given by:

$p_v =2*P(t_{9}>4.093)=0.00084$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

6 0

4 years ago