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miss Akunina [59]
3 years ago
7

Simplify the expression:3(7r + 1) =​

Mathematics
2 answers:
kozerog [31]3 years ago
8 0

Answer:

21r +3

Step-by-step explanation:

3(7r + 1) =​

Distribute

3*7r +3*1

21r +3

Andreas93 [3]3 years ago
4 0

Answer:

21r+3

Step-by-step explanation:

3(7r + 1)

Use the distributive property.

This property is normally used when the 2 numbers inside of the parentheses can not be added easily.

3*7r+3*1

21r+3

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Anyone know the answer to to this algebra question?
liraira [26]

Answer:

a=1

b= 1/16

c= 1/256

d= 1

e= 4/9

f= 16/81

Step-by-step explanation:

Plug in the values of X on the left side of the table into the function on the right side of the table. Remember that negative exponents mean the number is under a fraction. 4^-0 is like saying one over 4^0.

For the blue table, because the numbers are already in a fraction and in parentheses, you apply the exponent to each number individually.

7 0
3 years ago
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
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