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2 years ago

1 and the

Mathematics

1 answer:

umka2103 [35]2 years ago

6 0

Answer: There is a bunch of spelling mistakes in ur thing

Step-by-step explanation:

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Can someone please help me?

Lisa [10]

Answer:

Step-by-step explanation:

Using a graphing calculator, the graph shows an almost parabola like shape between 0 and 12

Plus if the width was equal or greater than twelve we would have a 0 or even negative volume which is impossible

7 0

2 years ago

Simplify (8m2+m+7m2)-(3m-4m3+2m2)

kondaur [170]

$(8m^2+m+7m^2)-(3m-4m^3+2m^2)\\\\=8m^2+m+7m^2-3m+4m^3-2m^2\qquad|\text{combine like terms}\\\\=(4m^3)+(8m^2+7m^2-2m^2)+(m-3m)\\\\=4m^3+13m^2-2m$

4 0

3 years ago

What is the product in lowest terms

Lostsunrise [7]

(8/15)*(-3/10)

multiply numerators and denominators: -24/150

simplify: -4/25

the answer is A.

5 0

3 years ago

What is -3.75 in the simplest form

NARA [144]

There is no simplest form for a decimal. You can only simplify equations and fractions, not decimals. For example, could you simplify 1? I don't think you could. Well, it's the same concept for -3.75.
Hope that answered your question.

5 0

2 years ago

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