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Phoenix [80]
4 years ago
8

X- 0.1x + 2.70 = 56.70 solve for the answer

Mathematics
1 answer:
Angelina_Jolie [31]4 years ago
5 0

Answer:

x=60

Step-by-step explanation:

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Convert 213base four to base5​
kakasveta [241]

Answer:

213 in base 4 = 124 in base 5

Step-by-step explanation:

First we convert base 4 to base 10

213 in base 4

2*42 + 1*41 + 3*40

32 + 4 + 3

39 in base 10

Now we convert base 10 to base 5

39/5  

7/5  r = 4

1 r = 2

124 in base 5

Hence 213 in base 4 = 124 in base 5

4 0
3 years ago
Find the equation of the line that passes through the points : (4 , -1 ) and (0 , -3)
Sergio [31]

Answer:

B. y= 1/2 x - 3

Step-by-step explanation:

base equation: y= mx+ b

slope(m) equation: \frac{y_{2}-y_{2}  }{x_{2} -x_{1} }

y2= -1    y1= -3

x2= 4     x1= 0

\frac{-1 - (-3)}{4-0} =\frac{2}{4} = \frac{1}{2}

y- intercept: -3

found by taking the y from (0,-3)

3 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
Which are factors of x2 – 4x – 5? Select two options. (x – 5) (x – 4) (x – 2) (x 1) (x 5).
Greeley [361]

Answer:

(x - 5) (x + 1)

Step-by-step explanation:

x² - 4x - 5

= x² + x - 5x - 5

= x(x + 1) - 5(x + 1)

= (x - 5) (x + 1)

5 0
2 years ago
Read 2 more answers
A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The prem
MAVERICK [17]

x = 160 gal

y = 80 gal

160 gallons of water and 80 gallons of premium antifreeze solution must be mixed.

8 0
3 years ago
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