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Eduardwww [97]
3 years ago
12

If the altitude of an isosceles right triangle has a length of x units, what is the length of one leg of the large right triangl

e in terms of x?
A) x units
B) x square root of 2 units
C) x square root of 3 units
D) 2x units

Mathematics
2 answers:
gulaghasi [49]3 years ago
6 0
Check the picture below.

bear in mind that, if the right-triangle is an isosceles, is a 45-45-90 triangle, and if we split it from the "vertex" above perpendicular to the base, we end up with two more 45-45-90 triangles.

const2013 [10]3 years ago
6 0

Answer:

B) x square root of 2 units

Step-by-step explanation:

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the answer to 54,000 x 10 is 540,000
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Find the difference: <br> -10 - (-6) = <br><br> A. -4<br><br> B. -16<br><br> C. +4<br><br> D. +16
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Answer:

-10-(-6) = -10+6 = -4; A

Step-by-step explanation:


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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:<em> if </em>f<em> is a continuous function on </em>[a,b]<em>, then</em>

                                   \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

3 0
3 years ago
Pls help (links will get reported)
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Answer:

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Step-by-step explanation:

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-1360+422= -938

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You first need to change the percent into a decimal then multiply by 80 to get your answer
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