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Murrr4er [49]
3 years ago
14

Can someone explain to me how I solve this problem. need íԵ as soon as possible. please help!❤️

Mathematics
1 answer:
densk [106]3 years ago
3 0
It says how many pizza was LEFT not how much they ate so there 13 pieces left over
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Please answer ASAP!!!!
trasher [3.6K]

Answer:

2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Which expression is equivalent to square root 55x^7y6/11x^11y^8
Novay_Z [31]

Answer:

\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}

Step-by-step explanation:

\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a

=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}

6 0
3 years ago
Find the arc length of the 3/4 of a circle with a radius of 5
IgorLugansk [536]

Answer:

7.5 pi

Step-by-step explanation:

The formula for arc length of a sector is denoted as

\frac{x}{360}2\pi r, where x is the central angle of the sector.

Since the sector is 3/4 of a circle, the central angle will be 3/4 of 360 degrees.

3/4 of 360 is 270, so we have our central angle. We also have our radius which we can plug into the formula.

\frac{270}{360}2(5)\pi

2 times 5 is equal to 10, and 270/360 simplifies to 3/4. 3/4 times 10 is equal to 7.5, so the answer is 7.5 pi

3 0
3 years ago
PLEASE HELP ME WITH THIS QUESTION!
RSB [31]
Peta has 10 plums. The second sentence is unnecessary.
8 0
3 years ago
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Help and explain explain !!!!!!!!!!
tigry1 [53]

Answer:

x=-1\text{ or }x=11

Step-by-step explanation:

For a=|b|, we have two cases:

\begin{cases}a=b,\\a=-b\end{cases}

Therefore, for 18=|15-3x|, we have the following cases:

\begin{cases}18=15-3x,\\18=-(15-3x)\end{cases}

Solving, we have:

\begin{cases}18=15-3x, -3x=3, x=\boxed{-1},\\18=-(15-3x), 18=-15+3x, 33=3x, x=\boxed{11}\end{cases}.

Therefore,

\implies \boxed{x=-1\text{ or }x=11}

3 0
2 years ago
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