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bulgar [2K]
3 years ago
9

Please help me it’s due today

Mathematics
1 answer:
-Dominant- [34]3 years ago
8 0
I might be wrong but probably is 25
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At a potato chip factory there were 92 machines working with each machine able to produce 56 chips a minute.If this is enough po
Mademuasel [1]
Divide 56 by 2 and you get 28,therefore 28 chips per box
3 0
3 years ago
find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller
Svetach [21]

Answer:

11, 13

Step-by-step explanation:

an odd number can be represented by 2n+1

since they are consecutive, the larger odd number will be 2n + 1 + 2

now,

2(2n+3) = 2n + 1 +15

solving this eqn, we get n = 5

so the two numbers are (2n+1) = 11 and 11+2= 13

4 0
3 years ago
Read 2 more answers
Can someone. Please help me please
zalisa [80]

Answer:

  • B. Always irrational

Step-by-step explanation:

The difference of the rational and and irrational number is always an irrational number.

Correct choice is B

5 0
3 years ago
Read 2 more answers
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
B2≤4<br><br> What is the solution to the inequality?<br><br> b≥8<br><br> b≤8<br> b≤2<br> b≥2
Allisa [31]
The solution is the third one:
b≤2
8 0
3 years ago
Read 2 more answers
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