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2 years ago

15/3 + (6,5 + 4,2) - 0,6 Please show your work, thank you :)

Mathematics

1 answer:

kifflom [539]2 years ago

3 0

Answer:

15/3 + (6.5 + 4.2) - 0.6 = 15.1

Step-by-step explanation:

You can first reduce the fraction to 5 when you multiply by 3, then calculate what's in the parenthesis. So now you have 5 + 10.7 - 0.6, then just solve from left to right. Hope this helps

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Steven's family needs a new microwave. There is one on sale that measures 20 inches by 14 inches by 11 inches. What is the volum

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Step-by-step explanation:

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3 years ago

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there are 2.54 centimeters in 1 inch. there are 100 centimeters in 1 meter.questionto the nearest meter, how many meters are in

Gre4nikov [31]

There are 7 meters in 279 inches.

There are 2.54 centimeters in 1 inch.
There are 100 centimeters in 1 meter.
The number of inches is given to be 279.
Unit conversion is a multi-step procedure that includes multiplying or dividing by a numerical factor, determining the appropriate number of significant digits, and rounding.
First of all, we need to convert inches to centimeters.
1 inch equals 2.54 centimeters.
279 inches equals 2.54*279 centimeters.
279 inches equals 708.66 centimeters.
Now, we need to convert these centimeters into meters.
Meters in 100 centimeters = 1
Meters in 1 centimeters = 1/100
Meters in 708.66 centimeters = (1/100)*708.66
Meters in 708.66 centimeters = 7.0866
Thus, there are approximately 7 meters in 279 inches.

To learn more about unit conversion, visit :

brainly.com/question/11543684

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1 year ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$