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3 years ago

7

The equation f = v + 80 represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,

overtime, t.
Which is an equivalent equation solved for t?
.
.
.
A) t = f - v/a
B) t = f - a/v
C) a(f - v)
D) v(f - a)

1 answer:

shusha [124]3 years ago

4 0

Answer:

A) t = (f -v)/a

Step-by-step explanation:

Undo the operations done to t, in reverse order. Here, t is multiplied by "a" then the product is added to v.

To solve for t, first you undo the addition of v by subtracting v from both sides of the equation.

f = v + at

f - v = at . . . . . subtract v

Then you undo the multiplication by "a" by dividing both sides of the equation by "a":

(f - v)/a = t . . . . . . sort of matches choice A

_____

Please note that the parentheses are required. f -v/a is something quite different. If this is typeset, the fraction bar does the grouping. If it is written in plain text, parentheses are required for grouping.

$t=\dfrac{f-v}{a}$

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2 years ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

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Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

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

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

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

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

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Take LCM

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$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

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