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3 years ago

I need the answer ? how can I solve the problem ?

Mathematics

1 answer:

Ilya [14]3 years ago

3 0

Answer:

use the standard equation of parabola.

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Which ratio is smaller 8 to 2 or 10 to 1

GalinKa [24]

Simplify the ratio 8 to 2 to be 4 to 1

Now compare 4 to 1 and 10 to 1

4 is smaller than 10 so the ratio 8 to 2 is smaller.

5 0

3 years ago

Use the diagram.<br> What is the length of }EG ?<br> A 1 B 4.4<br> C 10 D 16

nataly862011 [7]

Answer:

Option (D)

Step-by-step explanation:

Length of the bar EG = 1.6x

If F is a point on the bar such that,

EF + FG = EG

Measure of segment EF = 6

Measure of FG = x

By substituting measures of each side,

6 + x = 1.6x

1.6x - x = 6

0.6x = 6

x = 10

Length of EG = x + 6

= 10 + 6

= 16 units

Option (D) will be the correct option.

3 0

3 years ago

Factor completely 3a4y3 − 12a3y2 + 6a2y.

Firlakuza [10]

Step-by-step explanation:

$3a^4y^3-12a^3y^2+6a^2y=(3a^2y)(a^2y^2)-(3a^2y)(4ay)+(3a^2y)(2)\\\\=(3a^2y)(a^2y^2-4ay+2)$

5 0

3 years ago

A path starts and ends at diffrent verticles, and is allowed to repeat vertices and edges.

almond37 [142]

Answer:

Step-by-step explanation:

True

5 0

3 years ago

Read 2 more answers

The manufacturer of a CD player has found that the revenue R (in dollars) is Upper R (p )equals negative 5 p squared plus 1 com

AleksAgata [21]

Answer:

The maximum revenue is $1,20,125 that occurs when the unit price is $155.

Step-by-step explanation:

The revenue function is given as:

$R(p) = -5p^2 + 1550p$

where p is unit price in dollars.

First, we differentiate R(p) with respect to p, to get,

$\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550$

Equating the first derivative to zero, we get,

$\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155$

Again differentiation R(p), with respect to p, we get,

$\dfrac{d^2(R(p))}{dp^2} = -10$

At p = 155

$\dfrac{d^2(R(p))}{dp^2} < 0$

Thus by double derivative test, maxima occurs at p = 155 for R(p).

Thus, maximum revenue occurs when p = $155.

Maximum revenue

$R(155) = -5(155)^2 + 1550(155) = 120125$

Thus, maximum revenue is $120125 that occurs when the unit price is $155.

6 0

3 years ago