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Dovator [93]
3 years ago
5

How many ways can 8 different gifts be given to 5 diffrent children with each children recieving onegift?

Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0

8\cdot7\cdot6\cdot5\cdot4=6,720

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HELP! Write an equation that represents the relationship between the two variables.
vova2212 [387]

Answer:

x + 1, y + 1

Step-by-step explanation:

For every x there is a y

(っ◔◡◔)っ ♥ Hope It Helps ♥

7 0
2 years ago
Can someone solve this with steps cause idk how to solve the radicals
IceJOKER [234]
\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9


to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
8 0
3 years ago
Help with 1-9 , please!!!!
murzikaleks [220]
1. 160
You just add normally.
2. -86
3. 89
4. 7
5. -44
6. -14
7. -48
8. 50
9. 70
If you are struggling you can use a website called ixl it cost money but it can really help a lot !
4 0
3 years ago
Read 2 more answers
Given AB= BC<br> FINDIND <br> X=<br> AB=<br> BC=<br> AC=
tester [92]

Answer:

see explanation

Step-by-step explanation:

Given AB ≅ BC, then

AB = BC = x + 6 , and

AC = AB + BC ← substitute values

3x - 31 = x + 6 + x + 6, that is

3x - 31 = 2x + 12 ( subtract 2x from both sides )

x - 31 = 12 ( add 31 to both sides )

x = 43

Hence

AB = BC = x + 6 = 43 + 6 = 49

AC = 3x - 31 =(3 × 43) - 31 = 129 - 31 = 98

8 0
3 years ago
What are the restrictions for a? 2a^2+a-15/5a^2+16a+3
koban [17]

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


5a(a+3)+1(a+3)=0


We factor further to get,


(a+3)(5a+1)=0



This implies that,


(a+3)=0,(5a+1)=0


This gives


a=-3,a=-\frac{1}{5}


These are the restrictions.





8 0
3 years ago
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