All categories

3 years ago

Given the parent functions f(x) = log10 x and g(x) = 3x - 1, what is f(x) g(x)?

Mathematics

1 answer:

Rudiy273 years ago

7 0

Answer:

Option (b) is correct.

The value of $f(x)g(x)=log_{10}x(3x-1)$

Step-by-step explanation:

Given functions $f(x)=log_{10}x$ and $g(x)=3x-1$

We have to find $f(x)g(x)$

We know $f(x)g(x)=f(x) \cdot g(x)$

Thus, $f(x) \cdot g(x)=log_{10}x \cdot (3x-1)$

Thus, solving further we get,

$log_{10}x \cdot (3x-1)=log_{10}x(3x-1)$

Thus, the value of $f(x)g(x)=log_{10}x(3x-1)$

Thus, Option (b) is correct.

You might be interested in

Alaina is selling wreaths and poinsettias for her chorus fundraiser wreaths cost $27 each poinsettias cost $20 each if she sold

Olegator [25]

she sold 9 wreaths that day

3 0

3 years ago

Max was on vacation twice as long as Jared and half as long as Wesley. The boys were on vacation a total of 3 weeks. How many da

GuDViN [60]

Lets say W=Wesley, M=Max, and J=Jared.
So your equations are M=2J, 2M=W, and M+J+W=21days(3 weeks)
So substituting M and W into M+J+W=21 gives you 2J+J+2M=21, substitute M again, to get 2J+J+4J=21 so you get 7J=21 so J= 3 days. Using J=3, substitute in to find M, using M=2J, so M=2(3) so M=6 and using that, substitute into 2M=W, so 2(6)=W so W=12. And 12+6+3=21 to check.

6 0

3 years ago

Dr. Baker, a veterinarian, recorded the weight of each newborn kitten her office saw today.

Assoli18 [71]

There’s not actually a question. But I might be able to help if u show the full problem

7 0

2 years ago

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data

Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

The interval will be

$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0

3 years ago

What is an equation of the line that passes through the points (1, 6)<br> and (1, 2)?

victus00 [196]

Answer:y = -5x + 11

Step-by-step explanation:

First, find the slope (m) of the line that passes through these points:

m = (1-6) / (2-1) = -5/1 = -5

Then use the point slope formula to find the equation of the line:

m = (y-y0)/(x-x0)

-5 = (y-1)/(x-2)

-5(x-2) = y-1

y - 1 = -5x+10

y = -5x + 11

8 0

3 years ago