A^3-a^2-2a+8 factorize

F you have $18 and a 15% markup, what is the retail price?

Serggg [28]

OK so first I like to find what 10% is which in this case would be 1.8 and to find the other 5% you divide 1.8 by 2 which gets you .9 if you and 1.8 and .9 you get 2.7 and if you add 2.7 to 18 you get 20.7 which is your retail price

4 0

2 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

2 years ago

Find The Value Of Segment JK

kumpel [21]

Answer:

Step-by-step explanation:

since the two triangles are congruent,

GH = JK

Since, GH = 10 cm

JK is also 10 cm

Hope this helps

plz mark as brainliest!!!!!!

8 0

2 years ago

Simplify 4.2(2x-0.5y)-2(4.1x-0.3y)

klio [65]

Answer:

x = 7.5y

Step-by-step explanation:

Simplifying

4.2(2x + -0.5y) + -2(4.1x + -0.3y) = 0

(2x * 4.2 + -0.5y * 4.2) + -2(4.1x + -0.3y) = 0

(8.4x + -2.1y) + -2(4.1x + -0.3y) = 0

8.4x + -2.1y + (4.1x * -2 + -0.3y * -2) = 0

8.4x + -2.1y + (-8.2x + 0.6y) = 0

Reorder the terms:

8.4x + -8.2x + -2.1y + 0.6y = 0

Combine like terms: 8.4x + -8.2x = 0.2x

0.2x + -2.1y + 0.6y = 0

Combine like terms: -2.1y + 0.6y = -1.5y

0.2x + -1.5y = 0

Solving

0.2x + -1.5y = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '1.5y' to each side of the equation.

0.2x + -1.5y + 1.5y = 0 + 1.5y

Combine like terms: -1.5y + 1.5y = 0.0

0.2x + 0.0 = 0 + 1.5y

0.2x = 0 + 1.5y

Remove the zero:

0.2x = 1.5y

Divide each side by '0.2'.

x = 7.5y

Simplifying

x = 7.5y

5 0

3 years ago

A class consists of 10 male and 10 female students. if four students are selected at random, without replacement, what if four s

Flura [38]

$\displaystyle
|\Omega|=\binom{20}{4}=\dfrac{20!}{4!16!}=\dfrac{17\cdot18\cdot19\cdot20}{2\cdot3\cdot4}=4845\\
|A|=\binom{10}{4}=\dfrac{10!}{4!6!}=\dfrac{7\cdot8\cdot9\cdot10}{2\cdot3\cdot4}=210\\\\
P(A)=\dfrac{210}{4845}=\dfrac{14}{323}\approx4.3\%$

8 0

2 years ago

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A^3-a^2-2a+8 factorize​

A^3-a^2-2a+8 factorize