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Likurg_2 [28]
3 years ago
5

What expression is equivalent to 4(n- 3^2) + n? Thats 3 squared with 2

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

Given that,

An expression : 4(n- 3²) + n

To find,

The equivalent expression.

Solution,

We have,

4(n- 3²) + n

As 3²= 9

= 4(n- 9) + n

Opening the brackets,

= 4n-4(9) + n

= 4n-36 +n

= 5n-36

Hence, the equivalent expression is 5n-36.

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2n+5=20?????????????
Oksanka [162]

2n+5=20

Subtract 5 from 5 and 20

2n=15

Divide 15 and 2 by 2

n=7.5

:P

8 0
4 years ago
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What is the value of this expression? 6 (24÷ 3) - 2​
Makovka662 [10]

Answer:

6 (24÷3) -2

6 (8) -2

48-2

=46

Brackets of division multiplication add and subtract

5 0
3 years ago
9. Cal is buying T-shirts and shorts. T-shirts
damaskus [11]

Hey there!

Step By Step:

Well so we know that 12 is about half of an 25. So what we are gonna do is to buy 3 of 12$ and 2 of 25$.

3x12=36 and 2x25=50.

With adding those two last numbers, you’ll get $86.

Graph would be down below.

Hope it helps :D

3 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
The surface area of a rectangular prism that has height of 4 inches, width of 9 inches and length of 3 inches = ____________ in2
nevsk [136]

Answer:

150

Step-by-step explanation:

S.A.=2(wl+hl+hw)=

2((3x9)+(4x3)+(4x9))

2(27+12+36)

2(75)

150

8 0
3 years ago
Read 2 more answers
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