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RideAnS [48]
4 years ago
6

What's the answer???

Mathematics
1 answer:
mafiozo [28]4 years ago
5 0
I believe it is A. hope this helps
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Pls tell me the correct option
Hoochie [10]

Answer: I would go with C I hope i am right!!

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Mya and ERIC WORK IN A BAKERY . MYA HAS 3 TIMEA AS MUCH FLOUR AS ERIC .MYA HAS 15.81LB OF FLOUR .USE THE EQUATION 3F = 15.81 TO
Pavel [41]

Answer:

5.27 lb

Step-by-step explanation:

Given that,

Mya has 3 times as much floor as Eric. Mya has 15.81 lb of floor. The relation is given by :

Let Eric has F floor.

3F = 15.81 ...(1)

We need to find how much flour Eric has.

Dividing both sides by 15.81.

F = 15.81/3

F = 5.27 lb

Hence, Eric has 5.27 lb of flour.

4 0
3 years ago
Which inequality shows that Jerome wants to have read at least 85 pages by the end of Friday
Nikitich [7]

<em>Answer:</em>

<em>C</em>

<em>Hope this helps. Have a nice day.</em>

8 0
3 years ago
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PLSSSS I NEED HELPPPPP
Gelneren [198K]

Step-by-step explanation:

wjsixjq's with you and I have a nice day of my life with you to be a good time to do it in a couple days ago and I am looking to get back to the inbox to be a problem with this one and I have a nice day of school and work with you and the other side of things that I have a great day and the other side of my friends in the morning and will have the same to me that I have a great time in total there are any questions or if it is a great weekend as a few months I am going on with your phone is the best regards and thank u very much and have a great time with your phone number and I will be able and I will send you an

3 0
3 years ago
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Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
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