Pls help with this please

Mathematics

2 answers:

Marina86 [1]3 years ago

6 0

You have to do 2 5/8 4 times but in a picture or scrap

Nikitich [7]3 years ago

3 0

3. The answer is 5
4. The answer is 5
5. The answer is 16/3
6. The answer is 8
7. The answer is 3/2
8. The answer is 8
9.The answer is 6
10. The answer is 21/5
11. The answer is 10/3

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Explain your reasoning as you compare the values of a^n and a^-n when n<0 brainly

mojhsa [17]

n < 0, is another way to say "n is negative", so let's check

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad
a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
a^n~\hspace{10.5em}\stackrel{n = -n}{a^{-n}}\implies \cfrac{1}{a^n}
\\\\\\
a^{-n}~\hspace{10em}\stackrel{n=-n}{a^{-(-n)}}\implies a^{+n}\implies a^n$

4 0

3 years ago

Out of these triangles which of them is a Right Triangle?<br> 4, 6, 10<br> 6, 8, 10<br> 10, 24, 26

kupik [55]

The answer for you is B.
6, 8, 10

4 0

3 years ago

Halibut is $29 a pound .one portion is 6oz. How much does it cost to make one portion

natima [27]

It would be 10.87 because there is 16oz in a pound so 29/16= 1.8125 so 1.8125= 10.875

Hope this helps

Have a great day/night

8 0

3 years ago

A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$