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pickupchik [31]

2 years ago

12

A structure covering a sports practice field has the shape of half of a cylinder, as shown. What is the total surface area of th

e structure to the nearest square yard, including the ends and the flat area of the field itself? Use 3.14 for .

1 answer:

hoa [83]2 years ago

3 0

wait i dont know why it wont let me comment or see comments but wheres the picture?

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Please help me on this ASAP am I right ?

Paraphin [41]

Answer:

No, you would graph the y-intercept, you do not graph the m-value (slope). The slope value is the rate that the y-value increases per unit of x. This is how you graph your line from the y-intercept

8 0

2 years ago

Read 2 more answers

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$

4 0

2 years ago

AD is a common internal tangent to circles B and C. Find the length of the radius of circle B. Round to the nearest hundredth.

ycow [4]

ΔABE ~ ΔDCE
AB=AE*DC/DE=18*4/6=12

8 0

3 years ago

Read 2 more answers

Directions:Simplify the following monomials.SHOW ALL STEPS!

Mandarinka [93]

Answer:

<u>7. Ans</u>;

$\frac{6 {a}^{5} {b}^{7} }{ - 2 {a}^{3} {b}^{7} } = \frac{2 {a}^{3} {b}^{7}(3 {a}^{2} ) }{ - 2 {a}^{3} {b}^{7} } = - 3 {a}^{2}$

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<u>8. Ans</u>;

$\frac{ - 20 {x}^{3} {y}^{2} }{ - 5 {x}^{3}y } = \frac{ - 5 {x}^{3}y(4y) }{ - 5 {x}^{3}y } = 4y$

___o___o___

<u>9. Ans</u>;

$\frac{ - 16 a {b}^{4} }{4 {b}^{3} } = \frac{4 {b}^{3}( - 4ab) }{4 {b}^{3} } = - 4ab$

____o____o____

<u>10. Ans;</u>

$\frac{21 {m}^{8} {n}^{5} }{27 {m}^{5} {n}^{4} } = \frac{3 {m}^{5} {n}^{4}(7 {m}^{3} n) }{3 {m}^{5} {n}^{4}(9) } = \frac{7 {m}^{3}n }{9}$

___o___o___

<u>11. Ans;</u>

$\frac{ - 15 {x}^{5} {y}^{4} }{45x {y}^{3} } = \frac{ 15x {y}^{3} ( - {x}^{4} y)}{15x {y}^{3}(3) } = - \frac{ {x}^{4}y }{3}$

___o___o___

<u>12.Ans;</u>

$\frac{7 {p}^{2} {q}^{2} }{14 {p}^{2} {q}^{2} } = \frac{7 {p}^{2} {q}^{2} }{7 {p}^{2} {q}^{2} (2) } = \frac{1}{2} = 0.5$

I hope I helped you^_^

7 0

2 years ago

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Find the equation of the line<br> That is perpendicular to y=-2x+1 and contains the point (8,2)

skelet666 [1.2K]

Answer:

y=.5x-2

Step-by-step explanation:

If one line is perpendicular to another that means that their slopes are negative reciprocals

which means that the slope (or m in y=mx+b) is .5

so far we have

y=.5x+b

to solve for b we plug in (8,2)

2=.5(8)+b

2=4+b

-2=b

therefore our equation is

y=.5x-2

5 0

3 years ago

Read 2 more answers