How many numbers are equal to the sum of two odd one digit numbers

Given the function f(x) = 3x, find the value of f−1(81).

Shalnov [3]

Given the function f (x) = 3x, find the value of f-1 (81).
For this case, the first thing you should do is rewrite the function.
We have then:
y = 3 ^ x
From here, we clear the value of x:
log3 (y) = log3 (3 ^ x)
log3 (y) = x
Then, we rewrite the function again:
f (x) ^ - 1 = log3 (x)
Now, we evaluate the inverse function for x = 81:
f (81) ^ - 1 = log3 (81)
f (x) ^ - 1 = 4
Answer:
the value of f-1 (81) is:
f (x) ^ - 1 = 4

4 0

2 years ago

The length of side AI is?

marissa [1.9K]

Answer:

AI=3.25 IH= 4.2

Step-by-step explanation

The distance between C and D is 1.3 inches, the distance between E and F is 0.75 inches and the distance between G and H is 1.2 inches. This is true because the model says so. If you look closely together, this is equal to the distance between AI which is the length of AI. The answer would be 1.2+1.3+0.75=3.25. To find the length of side IH you do the same strategy, the distance between sides D and E is 4.8 inches, at the bottom is 9 inches of distance between side I and side F. 9-4.8 inches is equal to 4.2 inches. Therefore, the length of side AI is 3.25 inches and the length of side IH is 4.2 inches. If I am wrong please tell me for feedback, I also hoped that this has helped you in your learning :)

3 0

2 years ago

Read 2 more answers

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$