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2 years ago

14

Find the area of the polygon with the given vertices x(2,4) y(0,-2) z(2,-2)

1 answer:

weeeeeb [17]2 years ago

8 0

Answer:

6

Step-by-step explanation:

b = 2 units

h = 6 units

A=1/2 *b*h

= 1/2 *6*2

= 6

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Rodrigo made 6/8 of a pound of trail mix. If he puts 1/4 of a pound into each bag, how many bags can Rodrigo fill?

ValentinkaMS [17]

Answer:

He can fill 3 bags

Step-by-step explanation:

You would first make them have a common denominator so you would have 2/8 instead of 1/4 then you would do 6/8 divided by 2/8 to get 3

1/4 = 2/8

6/8 divided by 2/8 = 3

7 0

2 years ago

Read 2 more answers

A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

7 0

2 years ago

How do you solve for X ax-x=c

MArishka [77]

Factor out the "x" and you'll have:
$x(a-1) =c \qquad /:(a-1) \\ x=\frac{c}{a-1} \qquad \qquad a\neq 1$

8 0

2 years ago

Whats the value of x

Gnesinka [82]

Answer:

i think 64?

Step-by-step explanation:

its supposed to be the same on the bottom, right?

3 0

2 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