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Alexeev081 [22]
3 years ago
5

Lionel wanted to share 4 muffins with his 5 friends. He separated the muffins into 6 equal parts​

Mathematics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

24 muffins, each friend gets 4. There are 4 remaining

Step-by-step explanation:

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I’m stuck on this problem<br> Pls help ASAP
Viktor [21]

Answer:

x=+12

Step-by-step explanation:

4(x+6)=6x=16


We move all terms to the left:


4(x+6)-(6x)=0


We add all the numbers together, and all the variables


-6x+4(x+6)=0


We multiply parentheses


-6x+4x+24=0


We add all the numbers together, and all the variables


-2x+24=0


We move all terms containing x to the left, all other terms to the right


-2x=-24


x=-24/-2






4 0
3 years ago
Read 2 more answers
Write the equation of the line that passes through the points (3,4) and (1,5)
Illusion [34]
Hi,

y=  -\frac{1}{2} x+b



8 0
3 years ago
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What is the formula to a trapezoid
Molodets [167]
[(upper legth + lower length) x height] divided by 2
3 0
3 years ago
How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t
Zarrin [17]

By first principles, the derivative is

\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h

Use the binomial theorem to expand the numerator:

(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n

(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n

where

\dbinom nk=\dfrac{n!}{k!(n-k)!}

The first term is eliminated, and the limit is

\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h

A power of h in every term of the numerator cancels with h in the denominator:

\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)

Finally, each term containing h approaches 0 as h\to0, and the derivative is

y=x^n\implies y'=nx^{n-1}

4 0
3 years ago
A certain windshield wiper is 10 inches long. The rubber blade is 6 inches long and attached at the outside end. If the wiper sw
Sloan [31]
Area of sector=fraction of sector times area of circle
fraction of sector=degrees/360
and area of circle=pir^2

so
hold  a sec, we have the wiper on the outside, so we want to find the outside ring thing
do
area big sector-area small sector
or even better
(degrees/360) times (area big-areasmall)
big is 10
small is  4 (10-6=4)
areabig=10^2 times pi or 100pi
areasmall=4^2 times pi or 16pi
areabig-areasmall=100pi-16pi=84pi

then find the fraction

84pi times (150/360)=35pi square inches
4 0
3 years ago
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