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zhenek [66]
2 years ago
14

RATIO TABLES: An ostrich can run at a rate of 50 miles in 60 minutes. At this rate, how long would it take an ostrich to run 15

miles?
Mathematics
1 answer:
Marat540 [252]2 years ago
7 0
15      50
---- = ----
 x       60
(15*60)/50
900/50
18

It takes the ostrich 18 minutes to run 15 miles.
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Can some one please help me on my home work please i just need some one to show me how they solved this equation answers and wor
nignag [31]

Step-by-step explanation:

well first of all you have to expand the brackets. so 5(y+2)=4(x-3)

5y+10=4x-12

then you subtract 10

5y=4x-12-10=4x-22

then divide by 5 to get the value of y

y=(4x-22)/5

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2 years ago
Y= x +2<br> x^2 +y^2=10<br><br> solve using any method
Jobisdone [24]

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

(-3, -1) and (1, 3)

»»————- ★ ————-««  

Here’s why:  

  • I have graphed the two equations given on a graphing program.
  • When graphed, they pass at points (-3, -1) and (1,3). Therefore, they are the solutions to the system.

⸻⸻⸻⸻

See the graph attached.

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

7 0
3 years ago
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A roller coaster makes an angle of 55 degrees with the ground. The horizontal distance fro the crest of the hill to the bottom o
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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen
Jobisdone [24]

Parameterize S by the vector function

\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k

so that the normal vector to S is given by

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

with magnitude

\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}

In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

with magnitude \sqrt{1^2+2^2+1^2}=\sqrt6. The integral of f(x,y,z)=e^z over S is then

\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

where T is the region in the x,y plane over which S is defined. In this case, it's the triangle in the plane z=0 which we can capture with 0\le x\le8 and 0\le y\le\frac{8-x}2, so that we have

\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

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What’s the x your trying to find
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