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IRISSAK [1]
3 years ago
5

Mr. Burns built a wooden platform for the Drama Club at the school where he teaches. The shape below represents the dimensions o

f this wooden platform. (picture is not to scale) What is the surface area, in square feet, of this wooden platform?
​

Mathematics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

108 square feet

Step-by-step explanation:

First you want to break up the "L" into two parts. These parts can be seens as a and b. You need to find out the length on the left hand side, and half of it is 6ft as seen, so if half of it is 6ft, then you need to find the other half, which is obviously 6ft. 6+6=12, and to find the area of the thinner rectangle, multiply 2*24=48. That would be the area of section a. Section b is the bottom section and what you want to do is multiply 10*6 because this is how you would find the area of section b, which would be 60. 60+48=108. I know I was not very descriptive but I still hope this somewhat helps. I also might be wrong because of the fact that it says surface area despite the fact that this shape is 2D.

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Answer:

138g

Step-by-step explanation:

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4 0
3 years ago
1 х Given g(x)=
yarga [219]

(a) Since g(x)=\sqrt[3]{x} and h(x) = \frac1{x^3}, we have

(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x

We're given that

(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}

but we can rewrite this as

\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}

(bear in mind that we can only do this so long as <em>x</em> ≠ 0) so it follows that

f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}

(b) On its own, we may be tempted to conclude that the domain of (f\circ g\circ h)(x) = \frac1{1+x} is simply <em>x</em> ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.

g(x) = \sqrt[3]{x} is defined for all <em>x</em> - no issue here.

h(x) = \frac1{x^3} is defined for all <em>x</em> ≠ 0. Then (g\circ h)(x) = \frac1x also has a domain of <em>x</em> ≠ 0.

f(x) = \frac1{1+x} is defined for all <em>x</em> ≠ -1, but

(f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}

is undefined not only at <em>x</em> = -1, but also at <em>x</em> = 0. So the domain of (f\circ g\circ h)(x) is

\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}

7 0
3 years ago
A circle has a circumference of 14pi. What is the area in terms of pi?
Elza [17]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷The area in terms of pi is 49pi

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER

5 0
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