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4 years ago

How can complex fractions be used to solve problems involving ratios

1 answer:

5 0

When you have ratios, and some unknowns, you can create complex fractions from them. Bring them to the same denominator, and solve for x.

Example - we have this proportion:
2-x
5-45
And we can change it into fraction:
\frac{2}{5}=\frac{x}{45}

\frac{2*9}{5*9}=\frac{x}{45}

\frac{18}{45}=\frac{x}{45}
18=x

In case of more complex fractions it may come in handy.

You might be interested in

Which of the following functions is most closely connected with the FDA?

olga nikolaevna [1]

FDA stands for Food and Drug Administration.

Your answer is c.

Hope this helped.
(this question belongs in Health subject, not Mathematics.)

6 0

4 years ago

Read 2 more answers

find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B

SashulF [63]

Answer:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

$a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3$

$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

3 0

3 years ago

Answer and shown work for 5x + y = 10 -x - y = -2

konstantin123 [22]

Solve first equation for y:
5x+y=10
y=10-5x. Plug this equation into second equation

-x-(10-5x)=-2
-x-10+5x=-2
4x=8
x=2 (plug x into first equation)

5(2)+y=10
y=0

4 0

3 years ago

Read 2 more answers

QUICK I NEED HELP!!!

My name is Ann [436]

ITS B OR 92 DEGREES enjoy!!!

3 0

3 years ago

∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)

Nikolay [14]

keeping in mind that vertical angles are always congruent.

$\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}$

5 0

3 years ago