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3 years ago

Tom jogs 3 miles for every 8 miles that Sue jogs. Tom jogs 12 miles. How many miles does Sue jog?

Mathematics

1 answer:

RoseWind [281]3 years ago

8 0

Answer:

36 miles

The drawing will help

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Can someone please help me, I need step by step. Thank you

viktelen [127]

BCDE is a square of 576 square feet, so each side is √576=24 feet
BE=24
BF:FE=2:1, so BF=16, FE=8
Area of ABF is half of BE times AE (AE is the height): 1/2 of 16*12=96
the answer is 96 square feet

5 0

3 years ago

What is 3.141 divded by 0.45

kicyunya [14]

<span> 3.141 divided by 0.45 is 6.98.</span>

6 0

3 years ago

Read 2 more answers

Help please ty! (also ik theres alrdy an answer on the 2nd one but it was an accident :))

Umnica [9.8K]

The storage tank one is 14.99

Half cylinder is 471.239

I don't remember how to do the second one sorry :(

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3 years ago

Read 2 more answers

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

What is four more than six times a number is the same as nine times the number

svp [43]

Answer:

x = 1.33 I think.

Step-by-step explanation:

First you would write out your problem

Next you would take away 6 from both sides to get 4 = 3x

Now you would divide both sides by three to get x = 1.33

6 0

3 years ago