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3 years ago

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Choose the correct sum of the polynomials (6x3 − 8x − 5) + (3x3 + 6x + 2). 9x3 − 2x − 3 3x3 + 2x + 3 3x3 − 2x − 3 9x3 + 14x + 3

2 answers:

oksian1 [2.3K]3 years ago

7 0

Answer:

Option A. ( 9x³ - 2x - 3 )

Step-by-step explanation:

Given polynomials are ( 6x³ - 8x - 5 ) and ( 3x³ + 6x + 2)

We have to find the sum of these polynomials

( 6x³ - 8x - 5 ) + ( 3x³ + 6x + 2)

We will collect the similar terms first.

6x³ - 8x - 5 + 3x³ + 6x = 2

= ( 6x³ + 3x³ ) + ( 6x - 8x ) + ( 2 - 5 )

= ( 9x³ - 2x - 3 )

Option A. ( 9x³ - 2x - 3 ) is the correct option.

stiks02 [169]3 years ago

6 0

9x3 - 2x - 3
Simplify 6x3-8x-5 + 3x3+6x+2

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

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

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:

$z = 140 -x - y$

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

Take LCM

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

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

Hence, the numbers are:

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

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