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mario62 [17]
4 years ago
7

. Are the triangles similar and if so why? What is the value of x? Show your work. Answer:

Mathematics
1 answer:
alexdok [17]4 years ago
5 0

Answer:

  yes; x=7

Step-by-step explanation:

When two corresponding angles are congruent, all angles are congruent, so the triangles are similar. The sides of the triangle on the right are 30/25 = 6/5 times the length of the sides of the left triangle, so side MP will be 6/5 times the length of JL, or (6/5)·20 = 24.

Then the expression for that length lets us find the value of x:

  24 = 4x -4

  6 = x -1 . . . . . divide by 4

  7 = x . . . . . . . add 1

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The answer would be 7.225.
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3 years ago
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Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
3 years ago
Compare the functions shown below:
emmasim [6.3K]
The intersection with the y axis occurs when x = 0.
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f (0) = - 5

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h (0) = 2cos (0-\pi) +4

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3 years ago
The volume of the shape below is 6 square units. Determine the volume of the shape in two different ways.
REY [17]

The volume of the shape is 6 cubic units

<h3>How to determine the volume of the shape?</h3>

<u>Method 1</u>

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<u>Method 2</u>

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Read more about volumes at:

brainly.com/question/1972490

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