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tester [92]
3 years ago
10

(2/3)^3(1/5•2/3)^-3(1/5)^4 PLS HELP

Mathematics
1 answer:
Gnom [1K]3 years ago
4 0

Answer:

1/5

Step-by-step explanation:

You have to do whats in the parentheses first then you start doing to exponets from left to right then you would multiply then you would get your answer 1/5.

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The product of two positive integers is 120. One number is 2 less than five times the other. Select the equation that can be use
svetlana [45]

lets assume one number is x

then 2nd number will be 5x-2

so product is 120

x (5x-2) = 120

5x^2 - 2x = 120

so answer is last option

8 0
3 years ago
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What is the leading coefficient of the following polynomial?<br> 5x2-3x+6
gogolik [260]

Answer:

The 3

Step-by-step explanation:

3 0
3 years ago
After you add both sides by 9. How can you solve for 5z in this equation <br> 5z-3=12
lutik1710 [3]

Answer:

OPTION A: Divide both the sides by 5

Step-by-step explanation:

We are given:  5z - 3 = 12

When we add 9 to both sides we get:

5z - 3 + 9 = 12 + 9

⇒ 5z + 6 = 21

⇒ 5z = 21 - 6 = 15

⇒ 5z = 15

To find the value of z we have to divide both the sides by 5.

i.e., $ \frac{5z}{5} = \frac{15}{5} $

⇒ z = 3.

3 0
2 years ago
51/70 = x/100 Please answer I need help
BigorU [14]

Answer:

x=510/7

Step-by-step explanation:

51/70=x/100

cross product

70*x=51*100

70x=5100

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x=510/7

4 0
3 years ago
A. Compute the balance in an account if $25,000 is invested for 5 years with an APR of 3% compounded daily. The balance in the a
Sunny_sXe [5.5K]

Answer:

A.  $29,045.68

B.  $28,981.85

C. Compound daily applies interest everyday over a period of 365 days in a year. Compound annually only applies the 3% interest once a year.

Step-by-step explanation:

3% compounded daily:

   A = P(1 + r/n)^nt

      = 25000(1 + \frac{.03}{365})^365(5)

      = 25000(1.00008219)¹⁸²⁵

      = 25000(1.161827)

      = 29045.6770

3% compounded annually:

   A = P(1 + r/n)^nt

      = 25000(1 + \frac{.03}{1})^1(5)

      = 25000(1.03)⁵

      = 25000(1.159274)

      = 28981.8518

6 0
2 years ago
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