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3 years ago

2/7 - 5/7 Please Answer

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

8 0

Answer:

-5/7

Step-by-step explanation:

2-5 is -3 so the denominator stays the same so -3/7.

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Integral of sqrt(36+x^2)dx

AnnZ [28]

<span>Integral of sqrt(36 + x^2)dx = 1/2sqrt(36 + x^2)x + 18sinh^1 (x/6) + c </span>

4 0

3 years ago

Read 2 more answers

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>

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Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

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

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

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

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

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

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

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

4 0

3 years ago

Graph the solution x

mixas84 [53]

Answer:

I think you need to put more info

Step-by-step explanation:

8 0

3 years ago

What is the reason for statement 4 in this proof?

Naddika [18.5K]

It's the reflexive property because it's just relating it to itself (def. of reflexive)

As we all know: A=A, B=B, mx=mx, nothing changes.

So the answer is D

5 0

3 years ago

Read 2 more answers

Someone please help!!

rewona [7]

Answer:

Option 2: (1,0) is the correct answer

Step-by-step explanation:

Given inequality is:

y>-5x+3

In order to find which point is solution to the given inequality we'll put the point one by one in the inequality. If the point satisfies the inequality, then the point is the solution of the inequality.

Putting (0,3) in inequality

$3 > -5(0)+3\\3>0+3\\3>3$