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3 years ago

13

Floor is square with side length 12 feet. Kim wants the tiles to be squares with side length 2 feet. How many tiles does Kim nee

d to cover the entire floor.

1 answer:

Varvara68 [4.7K]3 years ago

6 0

Answer:

6 tiles

Step-by-step explanation:

Because 2,4,6,8,10,12

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30.000+100-5000*20=x(sum of profits)

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100 points please helpp

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Suppose you have a job teaching swimming lessons and get paid $6 an hour. You also have job as cashier and get paid $8 and hour.

VikaD [51]

This is a system of equations problem.

L = Hours of Swim Lessons
C = Hours as Cashier

L + C = 15 This is an equation for working 15 hours total.
so
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6 (15-C) + 8C = 100 *You substituted 15-C for L here
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L + C = 15
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L = 10 You can work 10 hours teaching swimming lessons.

If you want to make EXACTLY $100 you would work 5 hours as cashier and 10 hours teaching swimming lessons.

HOWEVER, the question says AT LEAST $100 and NOT MORE than 15 hours per week. Since you make more money as a cashier, any work over 5 hours will help you make over $100.

As long as you work at least 5 hours as a cashier and any remaining hours teaching swimming lessons, you will make over $100.

5 cashier, 10 swimming
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Mr. Pacey
JH/HS Social Studies Teacher
(but I've also helped with Math Team for JH & HS)

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2 years ago

Jonathan has 10 apples he takes away 9 how many does he have left

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Answer:

1

Step-by-step explanation:

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3 years ago

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Please help with this Calculus questions

Triss [41]

Answer:

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93$ .

Step-by-step explanation:

To find $\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du$ .

First, calculate the corresponding indefinite integral:

Integrate term by term:

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant rule $\int c\, du = c u$

$\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$

$2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$

$2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}$

Therefore,

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)$

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$ , so just evaluate the integral at the endpoints, and that's the answer.

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}$

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0$

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}$

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3 years ago