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3 years ago

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Last winter Armand had StartFraction 5 Over 6 EndFraction of a row of stacked logs. At the end of the winter, he had StartFracti

on 8 Over 15 EndFraction of the same row left. How much wood did he burn over the winter? 1 and StartFraction 9 Over 16 EndFraction rows 1 and StartFraction 11 Over 30 EndFraction rows StartFraction 4 Over 9 EndFraction row StartFraction 3 Over 10 EndFraction row

2 answers:

Nataly_w [17]3 years ago

6 0

Answer:

3/10

Step-by-step explanation:

According to the Question,

Given, Last winter Armand had 5/6 of a row of stacked logs & At the end of the winter he had 8/15 of the same row left.

Now, First We have to make the Fraction even so their denominators must have to be equal.

So. 15×2 = 30(denominator), Multiply with numerator too 8×2=16.

And, 6× 5 = 30(denominator), Multiply with numerator too 5×5=25.

Now. We get New Ratio's now with equal denominator, We can say that

⇒Armand had 25/30 of a row of stacked logs & At the end of the winter he had 16/30 of the same row left.

So, on Simply we get (25-16)/30 ⇒ 9/30 ⇒ 3/10 Wood he burns Over the winter.

hjlf3 years ago

5 0

Answer:

The answer is 3/10

Step-by-step explanation:

took the quiz recently and aced it with a 100 .

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3 years ago

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Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

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3 years ago