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2 years ago

Rearrange the formula F = mg + ma for a.

Mathematics

2 answers:

Lilit [14]2 years ago

8 0

Answer:

$\frac{F}{m}-g=a$

Step-by-step explanation:

We are given that a formula F=ma+mg

We have to rearrange the given formula for a.

The given formula can be written as

$F=m(a+g)$

Divide by m on both sides then we get

$\frac{F}{m}=a+g$

Subtract g on both sides then we get

$\frac{F}{m}-g=a+g-g$

$\frac{F}{m}-g=a$

Therefore, the formula for a is given by

$\frac{F}{m}-g=a$

Ivan2 years ago

4 0

draw out a triangle and then create three boxes inside by drawing a T. In each of the boxes you've created you need to put one of the letters from the equation. The equation you currently have is F= m*a. To rearrange this equation put the m and the a into the bottom boxes and the F above. Because there is a vertical line between the m and the a, this means you times them. If there is a horizonal line between two letters you divide them. So to find a, you must divide F by m.

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goldfiish [28.3K]

Answer:

0.05555555555

Step-by-step explanation:

usless you flip it around then it would be 2 cause 1/6 / 3 = .0555555555555 because the 1/6 butif they want u to flip it then it would be 6/3=2

6 0

2 years ago

Help please! I have choral practice...

Maslowich

This polygon is composed of a right triangle and a parallelogram.

The area of right angle :
$A_\Delta=\frac{1}{2}\cdot15\cdot8=\frac{1}{2}\cdot120=\boxed{60\ (cm^2)}$

The area of the parallelogram:
$A_P=15\cdot(13-8)=15\cdot5=\boxed{75\ (cm^2)}$

The area of the polygon is equal: $A=A_\Delta+A_P$

Therefore, the answer is:
$\boxed{\boxed{A=60\ cm^2+75\ cm^2=135\ cm^2}}$

7 0

3 years ago

Read 2 more answers

Someone please help me with this problem!!! I really need help and I don’t understand it!! If you know it please explain how you

nikitadnepr [17]

Distance = Rate * Time where the asterisk represents multiplication

We can isolate the rate if we divide both sides by the time. So we'll end up with this formula
Rate = Distance/Time
which can be written as
r = d/t

The distance is measured in feet while the time is measured in minutes. The rate r is measured in feet per minute. If the speed was 10 feet per minute, then this would mean that the object travels 10 feet every minute. Or put another way, when one minute passes by, the object has traveled 10 feet.

How did I get "feet per minute" from the equation r = d/t?
Well the units for "d" are in feet, and the units for "t" are in minutes

So writing "d/t" would turn into "feet/minute" when we think of the units only (ignore the actual numbers). Then we rewrite "feet/minute" into "feet per minute" as they mean the same thing, just written different ways.

-----------------------------------------------------

Answer: The units for the speed or rate are in feet per minute

3 0

3 years ago

What value of n makes the equation true? (2x^9y^n)(4x^2y^10)=8x^11y^20

Komok [63]

Hi there! The answer is n = 10.

$(2 {x}^{9} {y}^{n} )(4 {x}^{2} {y}^{10} ) = 8 {x}^{11} {y}^{20}$

As you see at the powers of x, we need to add the exponents of the power we when multiply them.
${x}^{9} \times {x}^{2} = {x}^{9 + 2} = x {}^{11}$

The powers of y work the same way.
${y}^{n} \times {y}^{10} = {y}^{n + 10} = {y}^{20}$

Hence, n = 10, since
${y}^{10 + 10} = {y}^{20}$

8 0

3 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers