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3 years ago

1/5 x 1/2 * mark you brainlyest

1 answer:

6 0

The answer will be 1/10. First you will change the denominator to the same number because you can’t mutiply fractions if the denominator is different. 5x2=10 and 2x5=10. So both of the denominator will turn into a 10. Now you’ll have to find the numerator. For the first one, multiply it by 2 because that how much the number need to turn equalvilant for the denominator. 1x2=2. So now you know that the first fraction is 2/10. Then you multiply 1 with 5 which is 5. Now the numbers all have the same denominator. Multiply 2/10 and 5/10 which is 1/10.

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Twenty reams of paper cost $47.20. How much does each ream of paper cost to the nearest cent?

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3 years ago

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Answer: x=2.375937.... sorry I give you the wrong one but here the steps O here x=2.96423 will it can be any one of the two answer I give you now.

Step-by-step explanation: Hope this help :D

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2 years ago

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Wittaler [7]

$4.30X40%=1.72
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2 years ago

1. The hypotenuse of a right triangle has a length of 10.0 and one of its

Mamont248 [21]

Answer:

I wont tell you the answer but I can tell you how to solve this one.

Step-by-step explanation:

If they are both squared then square them. Once you have the hypotenuse and the leg squared subtract the leg from the hypotenuse. That's your answer. Then you un- square it by finding the square root of the new number. That square root is your answer

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2 years ago

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.

6 0

2 years ago