Lola rode her bike for 7 of an hour on Saturday and of an hour on

What are the steps to (3w^3+7^2-4w+3) and (w+3)

Crazy boy [7]

Equation at the end of step 1 : Step 2 :Equation at the end of step 2 : Step 3 : 3w3 + 7w2 - 4w + 3 Simplify / w + 3 Checking for a perfect cube :

3.1 3w3 + 7w2 - 4w + 3 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 3w3 + 7w2 - 4w + 3

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -4w + 3
Group 2: 3w3 + 7w2

Pull out from each group separately :

Group 1: (-4w + 3) • (1) = (4w - 3) • (-1)
Group 2: (3w + 7) • (w2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(w) = 3w3 + 7w2 - 4w + 3
Polynomial Roots Calculator is a set of methods aimed at finding values of w for which F(w)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers w which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 3 and the Trailing Constant is 3.

The factor(s) are:

of the Leading Coefficient : 1,3
of the Trailing Constant : 1 ,3

Let us test ....

P Q P/Q F(P/Q) Divisor -1 1 -1.00 11.00 -1 3 -0.33 5.00 -3 1 -3.00 -3.00 1 1 1.00 9.00 1 3 0.33 2.56 3 1 3.00 135.00

Polynomial Roots Calculator found no rational roots

Polynomial Long Division :

3.4 Polynomial Long Division
Dividing : 3w3 + 7w2 - 4w + 3
 ("Dividend")
 By : w + 3 ("Divisor")

dividend 3w3 + 7w2 - 4w + 3 - divisor * 3w2 3w3 + 9w2 remainder - 2w2 - 4w + 3 - divisor * -2w1 - 2w2 - 6w remainder 2w + 3 - divisor * 2w0 2w + 6 remainder - 3

Quotient : 3w2 - 2w + 2
 Remainder : -3

Final result : 3w3 + 7w2 - 4w + 3 over w + 3

5 0

3 years ago

April rewrote a quadratic function in vertex form.

vivado [14]

Answer:

luke is starwars sun

Step-by-step explanation:

5 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

What is 2-2 equals to ?

Nikitich [7]

Answer:

The answer is 0

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Mr. Anderson wrote (7x9) 103 on the board. What is the value of that expression

RUDIKE [14]

Answer:

6,489

Step-by-step explanation:

first you would multiply 7 and 9, which would be 63

7 x 9 = 63

then you would multiply 63 by 103 since they are being multiplied

63 x 103 = 6,489

8 0

3 years ago