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Evgesh-ka [11]
3 years ago
8

Please help me with both problems thank you so much!!

Mathematics
2 answers:
motikmotik3 years ago
6 0
The one on the left is -5/6
The one on the right is 1/21
ohaa [14]3 years ago
5 0
E. -5/6
F. 1/21
That’s the answer
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a machine takes 1 minute to fill 6 cartons of eggs. at this rate how many minutes will it take to fill 420 cartons
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70 minutes, you divide 420 by 6

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4 years ago
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What is the solution to the system of equations below? y = 1/4x + 2 and 3y = - 3/4x -6
lord [1]

Answer:

<h2>x = -8 and y= 0 → (-8, 0)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}y=\dfrac{1}{4}x+2&(1)\\3y=-\dfrac{3}{4}x-6&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\3\left(\dfrac{1}{4}x+2\right)=-\dfrac{3}{4}x-6\qquad\text{use the distributive property}\\\\(3)\left(\dfrac{1}{4}x\right)+(3)(2)=-\dfrac{3}{4}x-6\\\\\dfrac{3}{4}x+6=-\dfrac{3}{4}x-6\qquad\text{subtract 6 from both sides}\\\\\dfrac{3}{4}x=-\dfrac{3}{4}x-12\qquad\text{add}\ \dfrac{3}{4}x\ \text{to both sides}\\\\\dfrac{6}{4}x=-12\qquad\text{multiply both sides by 4}\\\\6x=-48\qquad\text{divide both sides by 6}\\\\x=-8

\text{Put the value of x to (1):}\\\\y=\dfrac{1}{4}(-8)+2\\\\y=-2+2\\\\y=0

7 0
3 years ago
What is 70% from 70%
Nikitich [7]

The answer is 0.49. But if you meant what is 70% of 70 its 49

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3 years ago
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Guys, do anyone of you know how to do this? I don't know how! Please Help!!!!
Pavel [41]
A is the answer hope it helps
4 0
3 years ago
Why did I get this question wrong?
Vlad1618 [11]

You set up was almost accurate. Remember the arc length formula:

If f'(y) is continuous on the interval [a,b], then the length of the curve x = f(y), a ≤ y ≤ b should be;

L = ∫ᵇ ₐ √1 + [f'(y)]^2 * dy

We have to find the length of the curve given x = √y - 2y, and 1 ≤ y ≤ 4. You can tell your limits would be 1 to 4, and you are right on that part. But f'(y) would be rather...

f'(y) = 1/(2√y) - 2

So the integral would be:

∫⁴₁ √1 + (1/(2√y) - 2)² dy

Using a calculator we would receive the solution 5.832. Their is a definite curve, as represented below;

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