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3 years ago

Find the x- and y-intercepts of the function f(x) = log(2x + 1) − 1.

1 answer:

6 0

Alright, here's your answer.

y-intercept is computed (not found) by assigning x = 0 and computing y: here that is f(0) = Log(2*0 + 1) – 1 = Log(1) – 1 = 0 – 1 = -1
y-intercept is (0, -1)

x-intercept is computed by solving f(x) = 0 for x: here that is
0 = Log(2x + 1) – 1 → 1 = Log(2x + 1)

Assuming the Log cited is base 10, then 10^1 = 10^Log(2x + 1) = 2x + 1

That’s 10 = 2x + 1

Therefore 9 = 2x
x = 9/2 = 4.5
Check this result in the original equation, I did!

Your answer is - x-intercept is (4.5, 0)

I hope I helped! :)

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Multiply polynomials

Burka [1]

Answer -2x^4 - 2x^3 + 22x^2 - 30x + 12

Here's your solution

=> (2x^2 - 4x + 2) - (x^2 + 3x - 6)

=> (2x^2 - 4x + 2) -x^2 - 3x + 6

=> firstly multiple whole with -x^2

=> -x^2(2x^2 - 4x + 2) = -2x^4 + 4x^3 - 2x^2

=> now with -3x

=> -3x(2x^2 - 4x + 2) = -6x^3 +12x^2 -6x

=> with 6

=> 6(2x^2 - 4x + 2) = 12x^2 - 24x + 12

=> now add all terms

=> -2x^4 - 2x^3 + 22x^2 - 30x + 12

hope it helps and your day will be full of happiness ^_^

4 0

2 years ago

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ 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)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

3 0

3 years ago

SOMEONE WHO IS GOOD AT EASY MATH! (6th Grade Mathematics)

cricket20 [7]

Answer: Victims are naying that the rogue is a girl.

Step-by-step explanation:

4 0

2 years ago

Which of the following would be an acceptable first step in simplifying the expression sinx/1-sinx

Paha777 [63]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{sin(x)}{1-sin(x)}\implies \cfrac{sin(x)}{1-sin(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \stackrel{first~step}{\cfrac{sin(x)[1+sin(x)]}{[1-sin(x)][1+sin(x)]}}
\\\\\\
\cfrac{sin(x)[1+sin(x)]}{1^2-sin^2(x)}\implies \cfrac{sin(x)[1+sin(x)]}{cos^2(x)}
\\\\\\
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\\\\\\
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tan(x)[sec(x)+tan(x)]$

8 0

3 years ago

Read 2 more answers

What is the equation for the graph shown?

ELEN [110]

Your answer is c
Hope this helps :)

6 0

3 years ago