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4 years ago

14

You are arranging 7 books in a row on a shelf. one of the books is a dictionary and must go in the middle. how many ways can the

books be arranged?

1 answer:

alex41 [277]4 years ago

8 0

The other 6 books can be arranged in 6 ways

= 6*5*4*3*2*1 = 720 ways answer

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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>

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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

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

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

As

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

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

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

4 0

3 years ago

6x^2-5x+1<br><br> Factor polynomial

Tamiku [17]

6x²-5x+1
=6x²-3x-2x+1
=3x(2x-1)-1(2x-1)
=(3x-1)(2x-1)

8 0

3 years ago

Tyrell is a landscape architect. For his first public project, he is asked to create a small-scale drawing of a garden to be pla

dezoksy [38]

Answer:

D. is correct!

Step-by-step explanation:

ht tps://us-static.z-dn.net/fil es/ded/302e17a93f9f7dceb0b9635c302342ae.jp g

^ just an example

brainliest?

7 0

3 years ago

Please help ASAP!

Oliga [24]

The age of the students is about 0.67 years away from the 13.3 years.

The Mean Absolute Deviation is the average distance away from the mean. For the last choice, you should select more variable. Since the average is higher, there is a greater distance between the points and the mean.

3 0

3 years ago

Reduce fraction: a^3+a^2b/5a times 25/3b+3a

Margaret [11]

ANSWER

$\frac{5a}{3}$

EXPLANATION

The given fractions are:

$\frac{{a}^{3} + {a}^{2} b}{5a} \times \frac{25}{3b + 3a}$

We factor to obtain:

$\frac{{a}^{2}(a + b)}{5a} \times \frac{25}{3(a + b)}$

We cancel the common factors to get:

$\frac{{a}(1)}{1} \times \frac{5}{3(1)}$

We multiply the numerators and also multiply the denominators to get:

$\frac{5a}{3}$

Therefore the two fractions simplifies to $\frac{5a}{3}$

3 0

3 years ago