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3 years ago

The capacity at the local police 225 people

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

7 0

If they were 19& people at the pool what is the percentage 16.9

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Whenever I post a question nobody answer but when I put freebie everyone starts getting them in the second I post it like really

vlabodo [156]

Same tbh :( it sucks

8 0

3 years ago

Find dy/dx of y=csc(square root of x)

Vitek1552 [10]

Answer:

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$

Step-by-step explanation:

y = csc x

y' = -cot x csc x

$y = \csc \sqrt{x}$

$y' = \dfrac{d}{dx} [\csc \sqrt{x}]$

$y' = (-\cot x \csc x) \dfrac{d}{dx} \sqrt{x}$

$y' = (-\cot x \csc x) \dfrac{d}{dx} x^{\frac{1}{2}}$

$y' = (-\cot x \csc x) \dfrac{1}{2} x^{-\frac{1}{2}}$

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$

5 0

3 years ago

Please answer this correctly

natulia [17]

Answer:

Brainleist!

Step-by-step explanation:

10 marbles total

8 are either purple or orange

8/10 or 4/5 or 80% or 0.8

80 x 0.8 = 64

4 0

4 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>

First of all, I just would like to say:

$\text{Use } \LaTeX !$

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

Alex went bowling. He managed to get 3 strikes after 15 attempts. If you were to

algol13

Answer:

Theoretical Probability because its what should happen but experimental is what happens

Step-by-step explanation:

5 0

3 years ago