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Readme [11.4K]
2 years ago
5

Can someone please help me with this ​

Mathematics
2 answers:
earnstyle [38]2 years ago
8 0
The answer is C trust me
Elden [556K]2 years ago
3 0

Answer:

d

Step-by-step explanation:

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Select all the expressions that are equivalent to (2)^n+³
eimsori [14]

Answer:

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

  • a^{m} × a^{m}  = a^{m+n}
  • (a^{m})^{n} = a^{m*n}

Now let us find the equivalent expressions of  (2)^{n+3}

A.

∵ 4 = 2 × 2

∴ 4 =  2^{2}

∴  (4)^{n+2} =  (2^{2})^{n+2}

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴  (4)^{n+2} =  (2)^{2n+4}  

B.

∵ 4 = 2 × 2

∴ 4 =  2²

∴  4(2)^{n+1} = 2² ×  (2)^{n+1}

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴   4(2)^{n+1} =  (2)^{n+3}

C.

∵ 8 = 2 × 2 × 2

∴ 8 =  2³

∴  8(2)^{n} = 2³ ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴  8(2)^{n} =  (2)^{n+3}

D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = 2^{4}

∴  16(2)^{n} = 2^{4}  ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴  16(2)^{n} =  (2)^{n+4}

E.

(2)^{2n+3} is in its simplest form

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

3 0
3 years ago
Chanelle deposits $7,500 into the bank. She does not withdraw or deposit money for 6 years. She earns 6% interest during that ti
koban [17]

Answer: a. $2700

b. $10200

Step-by-step explanation:

a. The interest she would have earned at the end of the 6 years can be gotten using the formula

= PRT/100

= $7500 × 6% × 6

= $7500 × 6/100 × 6

= $7500 × 0.06 × 6

= $2700

b. Her balance when she wants to withdraw the money would be:

= $7500 + $2700

= $10200

5 0
3 years ago
Describe how to transform:
Anna71 [15]

Answer:

x^\frac{35}{6}

Step-by-step explanation:

The expression to transform is:

(\sqrt[6]{x^5})^7

Let's work first on the inside of the parenthesis.

Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:

\sqrt[n]{a} = a^{\frac{1}{n}}

Therefore \sqrt[6]{a} = a^{\frac{1}{6}}

Now let's replace a with x^{5} which is the algebraic form we are given inside the 6th root:

\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}

Now use the property that tells us how to proceed when we have  "exponent of an exponent":

(a^n)^m= a^{n*m}

Therefore we get:  (x^5)^{\frac{1}{6}}=x^{\frac{5}{6}}}

Finally remember that this expression was raised to the power 7, therefore:

[tex](\sqrt[6]{x^5})^7=(x^\frac{5}{6})^7=x^\frac{35}{6}[/tex]

An use again the property for the exponent of a exponent:

8 0
2 years ago
What is the least common denominator for the fractions 16and34?
fenix001 [56]

Answer:

4

Step-by-step explanation:

6 0
2 years ago
Two prime numbers that equal 75
Ber [7]
5 and 3 are two primary numbers for this question
4 0
3 years ago
Read 2 more answers
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