Simplify ( 3x ) -1.

Cha

attashe74 [19]

Answer:

3.1%

Step-by-step explanation:

6/196 × 100% = 3.1%

8 0

2 years ago

ANYONE...........help

KATRIN_1 [288]

Answer:

1 cm

Step-by-step explanation:

<u>Sides of the picture added margin:</u>

l = 8 + 2x
w = 5 + 2x

A(margin) = 30 cm²

<u>Solution</u>:

A(margin) = Total area - A(picture)
(8 + 2x)(5 + 2x) - 5*8 = 30
4x² + 10x + 16x + 40 - 40 - 30 = 0
4x² + 26x - 30 = 0
2x² + 13x - 15 = 0
x = (-13 + √(13² + 2*4*15))/4
x = (-13 + 17)/4
x = 1 cm

Note. The other root is ignored as negative.

3 0

3 years ago

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

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

5 0

3 years ago

Plzz help, will mark brainlest!

Alexus [3.1K]

☝that person in correct

8 0

2 years ago

A fully inflated basketball has a radius of 12

valentinak56 [21]

The volume of air needed is equal to the volume of the sphere, which is 7,234.56 cm^3.

<h3>How to get the volume of a sphere?</h3>

The volume of air that we need is equal to the volume of the basketball.

Remember that for a sphere of radius R, the volume is:

V = (4/3)*3.14*R^3

In this case, the radius is 12cm, replacing that we get:

V = (4/3)*3.14*(12cm)^3 = 7,234.56 cm^3

Then, to fully inflate the ball, we need 7,234.56 cm^3 of air.

If you want to learn more about spheres, you can read:

brainly.com/question/10171109

7 0

2 years ago