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seropon [69]
2 years ago
5

My last question someone help pls

Mathematics
1 answer:
lara [203]2 years ago
7 0

Answer:

3

Step-by-step explanation:

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Pleaseeeeeeeeeeee helpppppppppppppppp
gulaghasi [49]

Answer:

hirap naman yan basahin mo kasi directions

7 0
2 years ago
Find the surface area of the tool box. Round your answer to the nearest tenth and explain your answer. ​Pls I NEED THE ANSWER NO
Andrej [43]

Answer:

807.8 in^2

Step-by-step explanation:

The total area of the box is the sum of the areas of all faces of the box. The top, bottom, front, and back faces are rectangles 18 in long. The end faces each consist of a rectangle and a triangle. We can compute the sum of these like this:

The areas of top, bottom, front, and back add up to be 18 inches wide by the length that is the perimeter of the end: 2·5in +2·8 in + 9.6 in = 35.8 in. That lateral area is ...

(18 in)(35.6 in) = 640.8 in^2

The area of the triangle on each end is equivalent to the area of a rectangle half as high, so we can compute the area of each end as ...

(9.6 in)(8.7 in) = 83.52 in^2

Then the total area is the lateral area plus the area of the two ends:

640.8 in^2 + 2·83.52 in^2 = 807.84 in^2 ≈ 807.8 in^2

7 0
2 years ago
How to solve for the variable which is r
devlian [24]
I think, not 100% sure, but I think it is square root (A/4pi)=r
5 0
3 years ago
Read 2 more answers
Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)
viktelen [127]
\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}

The first line segment can be parameterized by \mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle with 0\le t\le1. Denote this first segment by C_1. Then

\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt
=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt
=108

The second line segment (C_2) can be described by \mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle, again with 0\le t\le1. Then

\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt
=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt
=-\dfrac{229}3

Finally,

\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3
5 0
3 years ago
Write 41/11 as a decimal
dimaraw [331]
41/11

Simply use your calculator

3.727272....
8 0
3 years ago
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