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Step2247 [10]
3 years ago
10

B + 12x

Mathematics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

Step-by-step explanation:

The expression shows the sum of two terms. The first term has coefficient 1. The second term is the product of 12 and x.

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Why do you need to calculate mean, or modo of a set of data?
goldfiish [28.3K]
To find the average of the data
3 0
3 years ago
Find the area 2 m 4 m 10 m
mina [271]

Answer:

30.285 m²

Step-by-step explanation:

C = 2πr

2 · π · 2

≈ 12.57

12.57 ÷ 2 = 6.285

10 - 2 = 8

8 · 4 ÷ 2

32 ÷ 2

16

2 · 4 = 8

6.285 + 8 + 16

30.285

6 0
3 years ago
Kamil invests 9000 for 4 years his investment pays compound interest of x% per anumn
elena-s [515]

To get the Total amount upon investment for the compound interest, plug in the value of x into the given expression bellow

<em>A = 9,000.00(1 +  x/100)^(4)</em>

Given data

Principal = $9000

Time = 4 years

Rate = x% per annum

<h3>Solution</h3>
  • Step one:

First, convert R as a percent to r as a decimal

r = x/100

r = x/100

  • Step two:

Then solve the equation for A

A = P(1 + x/100)^t

A = 9,000.00(1 + x/100)^(4)

A = 9,000.00(1 +  x/100)^(4)

  • Summary:

The total amount accrued, principal plus interest, with compound interest on a principal of $9,000.00 at a rate of x% per year.

Learn more about compound interest here:

brainly.com/question/24924853

6 0
2 years ago
A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t
nignag [31]
If A(t) is the amount of salt in the tank at time t, then the rate at which the amount of salt in the tank changes is given by

\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}
\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}

Let's drop the units for now. We have

\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24
e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}
\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}
e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt
e^{t/100}A(t)=2400e^{t/100}+C
A(t)=2400+Ce^{-t/100}

We're given that the water is pure at the start, so A(0)=0, giving

A(0)=0=2400+Ce^{-0/100}\implies C=-2400

So the amount of salt in the tank (in lbs) at time t is

A(t)=2400\left(1-e^{-t/100}\right)
4 0
3 years ago
In a certain mid-size city, 90% of the high school graduates are from public schools
Marat540 [252]

E.) .53

I just took this so I hope it helps

8 0
3 years ago
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