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3 years ago

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A toy factory makes 15 teddy bears every 10 minutes. The factory makes teddy bears for 8 hours each work day. What is the fewest

number of work days the factory will need to make 5,000 teddy bears? Write an explanation to show how you determined your answer

1 answer:

Kay [80]3 years ago

6 0

Ok so if they make 15 bears every 10 mins then they will make 90 bears in one hour.

15 x 6 = 90

If they work for 8 hours then they’ll make 720 bears in one work day.

90 x 8 = 720

Let x be the number of days...

720x = 5000

5000 / 720 = 6.9444444.....

Round up because they want the least number of work days

So the answer is 7 work days

Hope this helps! Any questions let me know :)

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If the ratio of side lengths of similar polygons is 6:11, what is the ratio of perimeters

algol13

Answer:

6 : 11

Step-by-step explanation:

the ratio 6 : 11 applies to all linear measure in the similar polygons

Both side lengths and perimeter are linear, hence

ratio of both is 6 : 11

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Answer:

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Step-by-step explanation:

y = mx (there is no intercept because you can't type anything in 0 minutes)

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4 years ago

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3 years ago

Read 2 more answers

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

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3 years ago

Write down in terms of n, an expression for the nth term of the following sequences 2,8,18,32,50,

Art [367]

Answer:

198

Step-by-step explanation:

2+6=8+10=18+14=32+18=50+22=72+26=96+30=126+34=160+38=198

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3 years ago