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3 years ago

5

Find the size of unknown anglese) 100°+a°

1 answer:

ANEK [815]3 years ago

8 0

Answer:

Ans=80

Step-by-step explanation:

You can see the picture for explanation.....

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Find the area bounded by the given curves: <br> y=2x−x2,y=2x−4

Andrej [43]

Answer:

$A = [\frac{32}{3}]$

Step-by-step explanation:

Given

$y_1 = 2x - x^2$

$y_2 = 2x - 4$

Required

Determine the area bounded by the curves

First, we need to determine their points of intersection

$2x - x^2 = 2x - 4$

Subtract 2x from both sides

$-x^2 = -4$

Multiply through by -1

$x^2 = 4$

Take square root of both sides

$x = 2$ or $x = -2$

This Area is then calculated as thus

$A = \int\limits^a_b {[y_1 - y_2]} \, dx$

<em>Where a = 2 and b = -2</em>

Substitute values for $y_1$ and $y_2$

$A = \int\limits^a_b {(2x - x^2) - (2x - 4)} \, dx$

Open Brackets

$A = \int\limits^a_b {2x - x^2 - 2x + 4} \, dx$

Collect Like Terms

$A = \int\limits^a_b {2x - 2x- x^2 + 4} \, dx$

$A = \int\limits^a_b {- x^2 + 4} \, dx$

Integrate

$A = [-\frac{x^{3}}{3} +4x](2,-2)$

$A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)]$

$A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}]$

$A = [\frac{16}{3}] - [\frac{-16}{3}]$

$A = [\frac{16}{3}] + [\frac{16}{3}]$

$A = [\frac{16 + 16}{3}]$

$A = [\frac{32}{3}]$

Hence, the Area is:

$A = [\frac{32}{3}]$

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4 years ago

Anyone wanna help me out :D (Brainly to correct one)

Yuliya22 [10]

Answer enclosed w/work

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3 years ago

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Dude help !!!!!!!!!!!

3241004551 [841]

it is option c, y=76x

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3 years ago

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I will make you Brainliest if you know what fraction is equal to 1.5

Igoryamba

Answer:

3/2

Step-by-step explanation:

There are a lot of fractions that equal 1.5, but if you're asking for the most simplified fraction, 3/2 is the answer.

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The line segment is extended to cross both axes. What is the location of the y-intercept on the line? O (0, -22)

riadik2000 [5.3K]

Answer: (0,-10)

Step-by-step explanation:

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