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maw [93]
3 years ago
10

What is the simplified form of the quantity x over 3 minus y over 4 all over the quantity x over 4 plus y over 3?

Mathematics
1 answer:
yaroslaw [1]3 years ago
3 0
First we write the expression:
 ((x / 3) - (y / 4)) / ((x / 4) + (y / 3))
 We multiply the numerator and the denominator of the expression by: (4) * (3).
 We have then:
 (((4) * (3)) * ((x / 3) - (y / 4))) / (((4) * (3)) * ((x / 4) + (y / 3)) )
 We now rewrite the expression:
 (4x - 3y) / (3x + 4y)
 The simplified expression is:
 (4x - 3y) / (3x + 4y)
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mestny [16]

Answer:

6*$10= $60

4*$10=$40

60+40= $100

$100

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is the factored form of this expression?<br><br> x2 − 121<br> HELP!!!!
olya-2409 [2.1K]

Answer:x= -11 or x= 11

Step-by-step explanation:tell me if you need more explaining

6 0
3 years ago
Roy and Sam start solving the equation as follows.
denpristay [2]

Answer:

The correct option is D.

Step-by-step explanation:

The given equation is

-3x+4=5x-6

According to the addition property of equality a=b and a+c=b+c are equivalent equations.

Use addition property of equality, add 3x on both the sides.

-3x+4+3x=5x-6+3x

4=8x-6

Therefore Sam's work is incorrect because he make calculation mistake.

According to the subtraction property of equality a=b and a-c=b-c are equivalent equations.

Use subtraction property of equality, subtract 5x from both the sides.

-3x+4-5x=5x-6-5x

-8x+4=-6

Therefore Roy's work is correct because he used subtraction property.

Option D is correct.

7 0
3 years ago
Jan saw 9 full moons in 252 days write each statement as a rate
Nataly [62]
I honestly don’t know sorry
6 0
2 years ago
How do you find a vector that is orthogonal to 5i + 12j ?
Rashid [163]
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\&#10;slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\&#10;-------------------------------\\\\

\bf \boxed{5i+12j}\implies &#10;\begin{array}{rllll}&#10;\ \textless \ 5&,&12\ \textgreater \ \\&#10;x&&y&#10;\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}&#10;\\\\\\&#10;slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}&#10;\\\\\\&#10;\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
4 0
3 years ago
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