Rewrite the equation in ax+by=c form. <br> Y+3=2(x-5)

11. The formula to find the AREA of a triangle is A = 1/2bh<br> True<br> False

Vera_Pavlovna [14]

Answer:

True

Step-by-step explanation:

The formula for the area of a triangle is:

area=(base x height)/2

7 0

3 years ago

The profit per acre from a grove of orange trees is given by x(190 − x) dollars, where x is the number of orange trees per acre.

nasty-shy [4]

Answer:

$P(x) = 190 x -x^2$

In order to maximize the last equation we can derivate the function in term of x and we got:

$\frac{dP}{dx} = 190 -2x$

And setting this derivate equal to 0 we got:

$\frac{dP}{dx} = 190 -2x=0$

And solving for x we got:

$x = 95$

And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

$P(x=95)= 95(190-95)= 95*95 = 9025$

Step-by-step explanation:

For this case we have the following function for the profit:

$P(x) = x(190-x)$

And we can rewrite this expression like this:

$P(x) = 190 x -x^2$

In order to maximize the last equation we can derivate the function in term of x and we got:

$\frac{dP}{dx} = 190 -2x$

And setting this derivate equal to 0 we got:

$\frac{dP}{dx} = 190 -2x=0$

And solving for x we got:

$x = 95$

And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

$P(x=95)= 95(190-95)= 95*95 = 9025$

7 0

3 years ago

Read 2 more answers

What is the vertex of the parabola y=2(x-3)^2+5 NEED ANSWER NOWWW

MArishka [77]

Answer:

(3,5)

Step-by-step explanation:

5 0

3 years ago

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$