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3 years ago

Does any one know how to solve this -4n^2-2n=-6-7n-5n^2

2 answers:

6 0

$-4n^2-2n=-6-7n-5n^2 \\ -4n^2-2n+6+7n+5n^2=0\\n^2+5n+6=0 \\ n^2+2n+3n+6=0 \\ n(n+2)+3(n+2)=0 \\ (n+2)(n+3)=0 \\ n+2=0 \ \ \ \ or \ \ \ \ n+3=0\\n=-2 \ \ \ \ \ \ \ \ \ \ \ \ \ n=-3$

Answer: n=-2, n=-3

zaharov [31]3 years ago

3 0

-4n^2-2n=-6-7n-5n^2

Add 5n^2 to both sides
n^2-2n=-6-7n

Add 7n to both sides
n^2+5n=-6

Add 6 to both sides to put the equation into quadratic standard form
n^2+5n+6=0

Factor the equation
(n+2)(n+3)=0

3+2=5, 3*2=6

To solve a quadratic equation, find roots using Zero Product Property.
(n+2)=0
n=-2

(n+3)=0
n=-3

Final answers: n=-3, n=-2

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The publisher of a recently released nonfiction book expects that over the first 20 months after its release, the monthly profit

wolverine [178]

Answer:

(a) $\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

(b)P'(5)=-($4.54) Thousand

(c)P'(11)=-($2.10) Thousand

(d)The fifth Month

Step-by-step explanation:

Given the monthly profit model:

$P(t)=\frac{240t-40t^2}{t^2+20}$

(a)We want to derive a model that gives the Marginal Profit, P' of the book.

We differentiate

$P(t)=\frac{240t-40t^2}{t^2+20}$ using quotient rule.

$\frac{dP}{dt}=\frac{(t^2+20)(240-80t)-(240t-40t^2)(2t)}{(t^2+20)^2}$

Simplifying

$\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

We have derived a model for the marginal profit.

(b) After 5 months, at t=5

Marginal Profit=P'(5)

$\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}$

$P^{'}(5)=\frac{4800-1600(5)-240(5)^2}{(5^2+20)^2}$

=-($4.54) Thousand of dollars

(c)Marginal Profit 11 Months after book release

$P^{'}(11)=\frac{4800-1600(11)-240(11)^2}{(11^2+20)^2}$

=-($2.10) Thousand of dollars

(d) Since the marginal profit at t=5 is negative, after the 5th Month, the profit starts to experience a steady decrease.

6 0

3 years ago

What is the format of this proof?

elena55 [62]

Divide b from both sides then it should be a equal C therefore if you add be it be the midpoint of a and C because BNB or equal to a C

4 0

3 years ago

What is the simplified expression for –3cd – d(2c – 4) – 4d?

kompoz [17]

–3cd – d(2c – 4) – 4d
= –3cd – 2cd + 4d – 4d
= -5cd

answer
A:–5cd

3 0

3 years ago

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andrew11 [14]

the correct answer should be c

5 0

4 years ago

Solve the system by substitution.

Butoxors [25]

Answer:

The solution set of given equations -x-y-z = -8 and - 4x + 4y + 5z = 7 and 2x + 2z = 4 is (3, 6, -1)

Solution:

Given, set linear equations are

-x – y – z = -8 ⇒ x + y + z = 8 → (1)

-4x + 4y + 5z = 7 ⇒ 4x – 4y – 5z = -7 → (2)

2x + 2z = 4 ⇒ x + z = 2 → (3)

We have to solve the above given equations using substitution method.

Now take (3), x + z = 2 ⇒ x = 2 – z

So substitute x value in (1)

(1) ⇒ (2 – z) + y + z = 8 ⇒ 2 + y + z – z = 8 ⇒ y + 0 = 8 – 2 ⇒ y = 6.

Now substitute x and y values in (2)

(2) ⇒ 4(2 – z) – 4(6) – 5z = - 7 ⇒ 8 – 4z – 24 – 5z = -7 ⇒ -9z – 16 = -7 ⇒ 9z = 7 – 16 ⇒ 9z = -9 ⇒ z = -1

Now substitute z value in (3)

(3) ⇒ x – 1 = 2 ⇒ x = 2 + 1 ⇒ x = 3

Hence, the solution set of given equations is (3, 6, -1).

4 0

3 years ago