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3 years ago

True or false -3>7(1)-10

Mathematics

2 answers:

Mice21 [21]3 years ago

5 0

Answer:

false because they are equal

Step-by-step explanation:

-3>7-10

-3>-3

false because they are equal

storchak [24]3 years ago

3 0

False, the equation is equal

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8 9/10-1 1/5 please answer soon

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Answer: 7 7/10

Step-by-step explanation:

First you have to multiply 1 1/5 by two so that you could subtract it from 8 9/10. Once you do that that’s when you get 7 7/10.

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

$\text{Use } \LaTeX !$

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

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KengaRu [80]

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Step-by-step explanation:

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anzhelika [568]

Answer:

150, 300, 450, 600, 750

Step-by-step explanation:

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