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harina [27]
3 years ago
9

I NEED HELP ASAP !!!!

Mathematics
1 answer:
mestny [16]3 years ago
8 0
D would me the answer I just took the assignment
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The escalator rises 25° from the ground and a vertical distance of 42 feet. How much horizontal distance does the escalator cove
zvonat [6]
Tangent (25) = opp / adj = 42 / horizontal distance
horizontal distance =  42 / tangent (25)
horizontal distance = 42 / 0.46631
horizontal distance = <span> <span> <span> 90.068838326435200 </span> </span> </span>
horizontal distance = 90.1 feet


5 0
3 years ago
[tex]\frac{3}{8} of 10x
pentagon [3]

Answer: x = 3/80

Step-by-step explanation:

5 0
2 years ago
greg is designing the clock face for a homemade clock. using the center of the clock face as the origin, he places the label 12
Serjik [45]
I apologized that you have not been helped yet, But the answer is (0,-5). Again, I am so sorry that you were not helped properly :)
8 0
3 years ago
Evaluate triple integral ∫ ∫ ∫ 8xydV, where E lies under the plane z = 1+x+y and above the E region in the xy-plane bounded by t
vazorg [7]

Answer:

\mathbf{=\dfrac{163.384}{15}}

Step-by-step explanation:

\int \int \limits_{E} \int \ 8 xy dV = \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} \int\limits^{1+x+y}_{0} \ 8xy dz dydx

= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} [ 8xyz]^{z=1+x+y}_{z=0}  \   \ dy dx

= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} 8xy (1+x+y) dy dx

= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} 8xy+8x^2y+8xy^2 \ \ dy dx

= \int\limits^{1}_{0}  \ [ 4xy^2+4x^2y^2+2.7xy^3]^{ y= \sqrt{x}}_{y-0} \ \  dx

= \int\limits^{1}_{0} \   4x (\sqrt{x})^2+4x^2(\sqrt{x})^2+2.7x(\sqrt{x})^3\ \  dx

= \int\limits^{1}_{0} \   4x^2+4x^3+2.7x^{5/2} \  dx

\mathbf{=\dfrac{163.384}{15}}

7 0
3 years ago
When point A( 3, 1) is translated 2 units to the right and 6 units up the coordinates of point A'
9966 [12]

Answer:

A' = (5,7)

Step-by-step explanation:

So, we start with the original point A at (3,1), that means x = 3 and y = 1.

We move it 2 units to the right... so we're moving along the X-axis... and we're moving to the right.. so we're increasing the value of x.  That means starting with x = 3, we go 2 more units... we then get to x = 5.

In the same way, we move 6 units up... so we move along the Y-axis.  We go up, so we also increase the value of y.  Starting with y = 1, we move up by 6 units... so we are now at y = 7.

A' point is then at (5,7).

3 0
2 years ago
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