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konstantin123 [22]

4 years ago

15

The following table shows the length and width of two rectangles:

1 answer:

faltersainse [42]4 years ago

5 0

The perimeter of these can be found by adding length and width and then multiplying by two.

Rectangle A: (with all of the variables already doubled)
2x + 16 + 2x - 2
2x + 2x + 16 - 2
4x + 14
So rectangle A's perimeter is 4x + 14.

Rectangle B: (Still with all the variables already doubled)
8x + 10 + 6x - 4
8x + 6x + 10 - 4
14x + 6
So rectangle B's perimeter is 14x + 6

And now to subtract the two.
(7x + 3) - (4x + 14)
14x + 6 - 4x - 14
14x - 4x + 6 - 14
10x - 8

So it would be C.

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4 years ago

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Bart works 36 hours a week and makes $612. Charles works 34 hours a week and makes $663. Who makes more per hour? How do you kno

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4 0

4 years ago

Involving the selection of frozen dinners. Assume that there are 15 frozen dinners: 7 pasta, 6 chicken, and 2 seafood dinners. T

zzz [600]

The probability that at least 2 of the dinners selected are pasta dinners will be 0.8181...

<u><em>Explanation</em></u>

Pasta dinners = 7 , Chicken dinners = 6 and Seafood dinners = 2

The student selects 5 of the total 15 dinners. So, total possible ways for selecting 5 dinners $=^1^5C_{5} =3003$

For selecting at least 2 of them as pasta dinners, the student can select 2, 3, 4 and 5 pasta dinners from total 7 pasta dinners.

So, the possible ways for selecting 2 pasta dinners $=^7C_{2}*^8C_{3} =1176$

The possible ways for selecting 3 pasta dinners $=^7C_{3}*^8C_{2} =980$

The possible ways for selecting 4 pasta dinners $=^7C_{4}*^8C_{1} =280$

The possible ways for selecting 5 pasta dinners $=^7C_{5}*^8C_{0} =21$

Thus, the probability for selecting at least 2 pasta dinners $=\frac{1176+980+280+21}{3003} =\frac{2457}{3003} =0.8181.....$

3 0

4 years ago

How do we do this out?

mars1129 [50]

$\dfrac{ {2x}^{2} + 7x + 6 }{x + 2} \\ \\ = \dfrac{(2x + 3)(x + 2)}{x + 2} \\ \\ = 2x + 3$

Answer : 2x + 3

Hope this helps. - M

6 0

4 years ago