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gizmo_the_mogwai [7]
3 years ago
7

Plsssss helppppppp because I am v confused​

Mathematics
1 answer:
Allisa [31]3 years ago
4 0
The simplified version of this expression is 110g^4h^16v^6. first, remove the parenthesis. the expression becomes 10g^3h^8v^6 times 11gh^8. then, calculate the product.
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Answer !! ( PLEASE SHOW WORK FOR NUMBER 5.) ( ZOOM IN TO SEE BETTER)
deff fn [24]

Answer:

12+2x

Step-by-step explanation:

2(10)+2(x-4)

20+2x-8

12+2x

5 0
3 years ago
PLEASE HELP ME WITH THANKS :)
Maurinko [17]

Answer

12855.04 ft³

Step-by-step explanation:

regular Octagon consists of 8 congruent triangles

B = (13.28 x 11) / 2 x 8=584.32

V = BH = 584.32 x 22 = 12855.04

8 0
3 years ago
Read 2 more answers
Is 3+5(9+2n)equal to48+5n
Rama09 [41]
Answer: No
There is no logical way to get ‘5n’
6 0
3 years ago
A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
If the cubic container had a volume of 1,728 in³, what would its dimensions be?
sineoko [7]

Answer:

12 inches × 12 inches × 12 inches

Step-by-step explanation:

Cube volume is s^3

s^3=1728

s=\sqrt[3]{1728}

s=12

6 0
3 years ago
Read 2 more answers
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