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3 years ago

I need to find the measure of each angle. I don't know where or how to start?

Mathematics

1 answer:

kirill115 [55]3 years ago

7 0

Answer:

40°

Step-by-step explanation:

Every triangle adds up to 180°

So to get the angle of the triangle we already have two values of:

$180 - 79 - 52 = 50$

Then since it's at a 90° angle we can subtract 50 to get:

40° as the other part of the right angle.

And applying the 180 rule again:

$180 - 100 - 40 = 40$

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What kind of triangle is 15,15,5

matrenka [14]

Answer:

isosceles triangle

Step-by-step explanation:

An isosceles triangle is a triangle with (at least) two equal sides. In the figure above, the two equal sides have length and the remaining side has length. . This property is equivalent to two angles of the triangle being equal. An isosceles triangle therefore has both two equal sides and two equal angles.

Brainiest??? plz

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4/5 times blank equals 2/5 what is blind

Sholpan [36]

If you would like to solve the equation 4/5 * x = 2/5, you can calculate this using the following steps:

4/5 * x = 2/5 /*5
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The correct result would be 1/2.

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3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$