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3 years ago

Find the surface area of the rectangular prism. 4 in 10 in 3 in

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

3 0

Answer: 120 in

Step-by-step explanation: You have to multiply all the numbers. Remember length.width.height is how you get your answer. .=multiply

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Rename the number 600,000 in ten thousands

telo118 [61]

I think it is 60,000

4 0

3 years ago

A circus tent is cylindrical upto a height of 3.3 m and conical above it if the diameter of the base is 100m and slant height of

Rina8888 [55]

Total cost of canvas required in making the tent is rupees 79128

Solution:

Given:

Height of the cylindrical tent = h = 3.3 m

Diameter of it's base = d = 100m

$radius = \frac{diameter}{2} = \frac{100}{2} = 50$

r = 50 m

Curved surface area of cylinder:

$C.S.A = 2 \pi rh$

$C.S.A = 2 \times 3.14 \times 50 \times 3.3 = 1036.2$

Thus Curved surface area of cylinder = 1036.2 square meter

Curved surface area of cone:

$C.S.A = \pi rl$

$C.S.A = 3.14 \times 50 \times 56.4 = 8854.8$

Thus Curved surface area of cone = 8854.8 square meter

Total curved area of tent:

⇒ curved area of cone + curved area of cylinder

$\rightarrow 8854.8 + 1036.2=9891$

Thus total curved area of tent = 9891 square meter

find total cost of canvas required in making the tent:

The cost of canvas is 8 rupees per square meter

$\rightarrow \text{total cost } = 9891 \times 8 = 79128$

Thus total cost of canvas required in making the tent is rupees 79128

7 0

3 years ago

Tommy is baking a cake. He knows that he needs 2/3 cups of flour to make 4 cakes. How many cups of flour will Tommy need to make

mash [69]

1. 1/3 cups..... ok vthen have a good day

4 0

3 years ago

PLEASE HELP ASAP!! (Find the surface area of the prism) SHOW WORK! (Number 6)

Leokris [45]

Sa= bw+hw+lw+ $\frac{1}{2}bh$ (2)
SA=(12*4)+(16*4)+(20*4)+( $\frac{1}{2} 12*16*2$ )
SA= 28+64+80+(6*16*2)
SA=28+64+80+(96*2)
SA=28+64+80+192
SA=364 $cm^{2}$

3 0

3 years ago

Analyze the asymptotes of the function.

koban [17]

Answer:

vertical asymptote at x=-1

horizontal asymptote at y=0

Step-by-step explanation:

$y=\frac{1}{x+1}$

To find vertical asymptote we set the denominator =0 and solve for x

x+1=0 (subtract 1 from both sides)

x=-1

So, vertical asymptote at x=-1

To find horizontal asymptote we look at the degree of both numerator and denominator

there is no variable at the numerator , so degree of numerator =0

degree of denominator =1

When the degree of numerator is less than the degree of denominator

then horizontal asymptote at y=0

5 0

3 years ago