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3 years ago

SOMEONE PLEASE HELP I WILL GIVE BRAINLIEST. (Use PEMDAS)

Mathematics

1 answer:

Slav-nsk [51]3 years ago

6 0

Answer:

n = 1/22

Step-by-step explanation:

[2/5 (8 + 2 · 6)] ÷ 2n = 88

2/5( 8 + 12) ÷ 2n = 88

2/5( 20) ÷ 2n = 88

8 ÷ 2n = 88

$\frac{4}{n}$ = 88

4 = 88n

n = 4/88 or 1/22

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When asked to find a number one-tenth as large as another number, what operation would you use? What about when asked to find a

kherson [118]

Apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

<u>Solution:</u>

Need to determine what operation is required to get one-tenth of a number and 10 times of a number

To get one tenth of a number, divide the number by 10.

For example to get one – tenth of 100, divide it by 10, we get 10 as a result.

$\frac{1}{10} \times 100 = 10$

To get ten times of a number, multiply the number by 10

For example 10 times of 10 = 10 x 10 = 100

Hence apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

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3 years ago

Evaluate the expression. 8 + 12 • 7

arsen [322]

Multiply first, then add. 12 times 7 = 84. Then add 8. 84 + 8 = 92.

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4 years ago

Read 2 more answers

(5 - 3)⁴ - 2(7) + 8² = ?

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

(5 - 3)⁴ - 2(7) + 8²

PEMDAS

Parentheses first

(2)⁴ - 2(7) + 8²

Exponents

16 - 2(7) + 64

Multiply and divide from left to right

16 -14 +64

Add and subtract from left to right

2+64

7 0

3 years ago

A circle has a diameter with endpoints at –2 + i and –6 –11i. What is the center of the circle?

kotykmax [81]

Be more specific. Like what is it asking you to do ? I'm trying to help. :)

4 0

4 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago