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3 years ago

Solve the equation. Show your work.6x-15=3(4x+3)

2 answers:

7 0

6x-15=3(4x+3)
Distribute the three to the parentheses
6x-15=12x+9
Transfer the 6 over to the 12 by subtracting it
-15=6x+9
Subtract 9 from 15
-24=6x
Then divide the six by the -24
X=-4

Law Incorporation [45]3 years ago

4 0

$6x - 15 = 3(4x + 3)$
then you distribute the three
so you would do
$3 \times 4x = 12x$
$3 \times 3 = 9$
$6x - 15 = 12x + 9$
then you get the variables on the same side
$- 6x$
on each side so you would have 6x on the right side then you subtract the nine so you would have -24 on the left side so now you have
$- 24 = 6x$
then you divided by 6 which the answer equals
$x = - 4$

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The demand function for q units of a product at $p per unit is given by p(q + 3)2 = 100,000. Find the rate of change of quantity

wariber [46]

Answer:

-0.625 unit

Step-by-step explanation:

Given that:

$p(q+3)^2 = 100, 000$

where;

p = price

q = quantity

To find the rate of change of quantity (q) with respect to price (p) we go by the differentiation

$(q+3)^2\frac{dp}{dp} +p^2(q+3)\frac{dq}{dp}=0$

$(q+3)^2 +p^2(q+3)\frac{dq}{dp}=0$

$\frac{dq}{dp}=\frac{-(q+3)}{2p}$

when P =40

Then 4(q+3)² = 100, 000

(q+3)² = $\frac{100,000}{4}$

(q+3)² = 25,000

(q+3) = $\sqrt{2500}$

q+ 3 = 50

q = 50 -3

q = 47

NOW; $\frac{dq}{dp}=\frac{-(q+3)}{2p}$

$\frac{dq}{dp}=\frac{-(47+3)}{2*40}$

$\frac{dq}{dp}=\frac{-47-3}{2*40}$

$\frac{dq}{dp}=\frac{-50}{80}$

$\frac{dq}{dp}=-0.625$

Thus, the rate of change of quantity with respect to price when p = $40 is -0.625 unit.

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4 years ago

A square pyramid has a base 5 meters in length and a height of 5 meters. Approximate the volume of the object.

Ierofanga [76]

Final Answer: 41.7

The other person who answered is incorrect. I completed an assignment just now with the exact question listed above, and my answer (41.7) was correct.

You have to use this equation:

V = 1/3 (5 • 5) (5)

V = 1/3 • 25 • 5

V = 125/3

V = 41.6667

V = 41.7 (rounded to nearest hundreth)

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