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Stolb23 [73]
3 years ago
9

What is the solution to w-9 1/2=15 A)5 1/2 B 6 1/2 C) 241/2 D) 23 1/2

Mathematics
2 answers:
erik [133]3 years ago
8 0
The answer is C. 241/2
cupoosta [38]3 years ago
4 0
The answer is C. 24 1/2
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3 years ago
How to solve: a parade traveled 4 miles in 2 hours. how far did tge parade travel per minute?
jek_recluse [69]

There's a really easy way to convert any units to other units.

Right now, we have the fraction  (4 miles) / (2 hours).

We want to find a fraction that's exactly equal to that one,
but has the units of  (miles/minute)  or maybe  (feet/minute).

Just take the original fraction, and multiply it by some other
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Each fraction you multiply it by must have the value of ' 1 ' so
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The second fraction is equal to ' 1 ', because the top and the bottom
have the same value ...  1 hour is the same thing as  60 minutes.

Multiply the fractions:  (4 miles x 1 hour) / (2 hour x 60 minutes)

Now you can cancel 'hour' from the top and the bottom, and you have

             (4 miles x 1) / (2 x 60 minutes)

               = (4 miles) / (120 minutes) 

               =          (4 / 120) mile/minute = 0.0333... mile / minute .

Let's do it again, go a little farther, and get an answer that
might mean more and feel more like an answer. 

   (4 miles) / (2 hours) x (5280 feet / mile) x (1 hour / 60 minutes)

The 2nd and 3rd fractions both have the value of ' 1 ', because
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Multiply all three fractions: 

     (4 miles x 5280 feet x 1 hour) / (2 hours x 1 mile x 60 minutes)

You can cancel both 'mile' and 'hour' out of the top and bottom,
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     (4 x 5280 feet x 1) / (2 x 1 x 60 minutes)

  =  (4 x 5280) / (2 x 60)  feet / minutes

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3 0
2 years ago
Read 2 more answers
Find the work done by force dield f(x,y)=x^2i+ye^xj on a particle that moves along parabola x=y^2+1 from(1,0) to (2,1)
NemiM [27]
Call the parabola P, parameterized by \mathbf r(y)=\langle y^2+1,y\rangle with 0\le y\le 1\rangle. Then the work done by \mathbf f(x,y) along P is

\displaystyle\int_P\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{y=0}^{y=1}\mathbf f(x(y),y)\cdot\dfrac{\mathrm d\mathbf r(y)}{\mathrm dy}\,\mathrm dy
=\displaystyle\int_0^1\langle(y^2+1)^2,ye^{y^2+1}\rangle\cdot\langle2y,1\rangle\,\mathrm dy
=\displaystyle\int_0^1(2y^5+4y^3+2y+ye^{y^2+1})\,\mathrm dy=\frac73-\frac e2+\frac{e^2}2
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3 years ago
A student weighs a sample of the shale and
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interesting choice of pfp

4 0
2 years ago
The triangle below is equilateral. Find the length of side x to the nearest tenth.
Gemiola [76]

Answer:

\sqrt{\frac{15}{2} } or 2.738

Step-by-step explanation:

Let’s just look at the triangle on the top with the \sqrt{10} on the top and x on the bottom. (Basically the top half to the equilateral triangle)

There is a small square in the bottom right corner, which indicates that this triangle is a right triangle. This means that we can use the Pythagorean Theorem: a^{2} +b^{2} =c^{2}

We know that \sqrt{10} is our hypotenuse, and therefore our c in our equation. Let’s say that x=a in our equation. Therefore we are left to find b. However, b is half the length of the side of the original equilateral triangle. An equilateral triangle means that all three sides are the same length. Therefore our side would also be \sqrt{10} units long. However we know that b is half of that value, so b=\frac{1}{2}(\sqrt{10}) or \frac{\sqrt{10} }{2}

Plugging these values into the equation:

x^2+ (\frac{\sqrt{10} }{2})^{2}=\sqrt{10} ^{2}

x^{2}=\sqrt{10} ^{2}-\frac{\sqrt{10} }{2} ^{2}

x^{2} =10-\frac{10}{4}

x^{2} =\frac{15}{2}

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This approximately equals 2.738

5 0
1 year ago
Read 2 more answers
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