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Shtirlitz [24]
2 years ago
14

Simplify ( 3x ) -1.

Mathematics
1 answer:
klasskru [66]2 years ago
3 0

Answer:

B

Step-by-step explanation:

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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
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