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topjm [15]
3 years ago
13

Please and thank you​

Mathematics
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

.3125

Step-by-step explanation:

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Help pls
timurjin [86]

Answer:

s> 144 2/3

Step-by-step explanation:

I took the test in k12 and i got it right so yea !

3 0
2 years ago
Solve the equation 1/6(x - 5) = 1/2(x + 6)
Semmy [17]

Answer:

-23/2

Step-by-step explanation:

x-5=3(x+6)

x-3x=18+5

-2x=23

x=-23/2

5 0
3 years ago
Convert 2000 g/l to the unit g/ml
balu736 [363]
\frac{2 \000 g}{l} = \ ? \ \frac{g}{ml} \\ \\ 1 \ l = 1 \ 000 \ ml \\ \\\frac{2\ 000 \ g}{l} =\frac{2 \000 \ g}{1 \000\ ml} = 2.0 \ \ \frac{g}{ml}


6 0
3 years ago
Read 2 more answers
The Math Club had a car wash to raise money for a competition. The members charged $10 for each car they washed. What is the ind
Fudgin [204]

Answer:

The dependent variable for this case is the amount of money charged and the independent variable would be the number of cars washed.

And the reason why is because the dependnet variable is the variable of interest (money earned) and the independent the variable that controls the amount of money earned

Step-by-step explanation:

From the info given by the problem we know that the Math Club had a car wash to raise money for a competition. The members charged $10 for each car they washed.

The dependent variable for this case is the amount of money charged and the independent variable would be the number of cars washed.

And the reason why is because the dependnet variable is the variable of interest (money earned) and the independent the variable that controls the amount of money earned

8 0
3 years ago
What is the area of the region bounded between the curves y=x and y=sqrt(x)?
erastova [34]

Answer:

\frac{1}{6}

Step-by-step explanation:

y = x     .....(1)

y=\sqrt{x}     .... (2)

By solving equation (1) and equation (2)

x = \sqrt{x}

\sqrt{x}\left ( \sqrt{x}-1 \right )=0

\sqrt{x}=0 or \left ( \sqrt{x}-1 \right )=0

x = 0, x = 1

y = 0, y = 1

A = \int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)

A = \int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx

A = \frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}

A = \frac{2}{3}-\frac{1}{2}

A = \frac{1}{6}

8 0
3 years ago
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