AC=5x-4 and CD=16-3x solve for x

If SDE ~ SWT, find WT.

faust18 [17]

Formula used: 40+x = 4x-3 x is used for ET

Step-by-step explanation:

40+x = 4x-3

+3 +3

43+x = 4x

-x -x

43 = 3x

3 3

14.3 = x

then substitude this on 5x+3 on WT and you arr good

5 0

2 years ago

Find the values of x and y. 48° (5x – 17) X = and y =

likoan [24]

Can you explain more

6 0

2 years ago

Simplify the expression 2(x-6)+2(8)

jolli1 [7]

Answer:

2x+4

Step-by-step explanation:

Distribute:

=(2)(x)+(2)(−6)+(2)(8)

=2x+−12+16

Combine Like Terms:

=2x+−12+16

=(2x)+(−12+16)

=2x+4

8 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 a separable differential equation. 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.

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