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MaRussiya [10]
3 years ago
7

Simplify this expression 6w+4+8w-1

Mathematics
2 answers:
FrozenT [24]3 years ago
6 0

Answer:

14w + 3

Step-by-step explanation:

6w + 4 + 8w -1 (Given)

14w + 3 (combine like terms)

<u><em>"The Kid Laroi" - Rapper-Songwriter.</em></u>

den301095 [7]3 years ago
5 0

Answer:

14w+3

Step-by-step explanation:

6w+4+8w-1

(6w+8w)+(4-1)

14w+3

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Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol
koban [17]

Answer:

\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}

Step-by-step explanation:

\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228

=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}

\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}

4 0
2 years ago
What is the value of Fraction 1 over 2x3 + 3.4y when x = 4 and y = 2?
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Plug in x and y

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3 years ago
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3 years ago
Shane purchased a television at a 20% discount. If the sale price is $375.00, what is the original price ?​
Sveta_85 [38]

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2 years ago
Evaluate the expression for the given value of the variable.<br><br><br>n=500<br>∛n/4+n/10=?
schepotkina [342]

Answer:

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Step-by-step explanation:

To solve the expression, \sqrt[3]{\frac{n}{4}} + \frac{n}{10} substitute n = 500 and simplify according to order of operations.

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2 years ago
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