$1.\:2^x=64, x=\boxed{6},\\2.\:x=\left(\frac{2}{3}\right)^3,x=\boxed{\frac{8}{125}}\\3.\:3\cdot (3^4)=3^x, x=\boxed{5}\\4.\:\frac{16}{25}=x^2,x=\boxed{\frac{4}{5}},$

Step-by-step explanation:

$1.\\\\2^x=64,\\\log 2^x=\log64,\\x\log 2=\log 64,\\x=\frac{\log 64}{\log 2}=\boxed{6}$

$2.\\x=\left(\frac{2}{5}\right)^3=\frac{2^3}{5^3}=\boxed{\frac{8}{125}}$

3. This problem incorporates an exponent property.

Exponent property used: $a^b\cdot a^c=a^{(b+c)}$ , yielding an answer of $\boxed{5}$

$4.\\\\\frac{16}{25}=x^2,\\x=\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\boxed{\frac{4}{5}}$

3 0

3 years ago

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Given the Arithmetic sequence A1,A2,A3,A4 53, 62, 71, 80 What is the value of A38?

Murrr4er [49]

Answer:

$A_{38} = 350$

Step-by-step explanation:

The 5th term of the arithmetic sequence is 53. We can write the equation:

$a + 4d = 53...(1)$

The 6th term of the arithmetic sequence is 62. We can write the equation:

$a + 5d = 62...(2)$

Subtract the first equation from the second one to get:

$5d - 4d = 62 - 53$

$d = 9$

The first term is

$a + 4(9) = 53$

$a + 36 = 53$

$a = 53 - 36$

$a = 17$

The 38th term of the sequence is given by:

$A_{38} = a + 37d$

$A_{38} = 17+ 37(9)$

$A_{38} = 350$

3 0

3 years ago

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Identify the fractions labeled with the letters A and B on the scale.

romanna [79]

A is 1/8 and b is 5/8, hope this helps!

7 0

3 years ago

What is an equation of the line that passes through the points (3,4) and (5,8)

algol [13]

Answer:

i think 5

Step-by-step explanation:

6 0

3 years ago

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