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zavuch27 [327]
1 year ago
7

Which table is a nonlinear function?

Mathematics
1 answer:
lisov135 [29]1 year ago
6 0

Answer:

example to solve it

Step-by-step explanation:

here is a postulate (a rule) from geometry that says any two points determine a line. 1. Pick two points from the above set, for example (7,7) and (1,9). 2. Calculate the slope of the line between them: m = (9 - 7)/ (1 -7)= - 2/6 = -1/3. 3.

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Jim had 5/7 of a pound of candy. she ate 1 3/4 of that candy. how many pounds of candy did he eat?
harkovskaia [24]
What she ate is
(13/4)×(5/7)
65/28 pounfs of candy
5 0
3 years ago
What is the slope of the line that passes through the points (8,2) and (10,2)? Write your answer in the simplest form.​
Oxana [17]

slope of line=(y2-y1)/(x2-x1)

give is

(x1,y1)=(8,2)

(x2,y2)=(10,2)

therefore slope =(2-2)/(10-8)=0/2=0

Hence

answer

Slope of line=0

3 0
2 years ago
14^-3x14^5 using a single positive exponent
lesya692 [45]

Answer:

1/14^3x times 14^5

Step-by-step explanation:

Sorry I dont have one but I hope this helps, GodBless.

6 0
3 years ago
Read 2 more answers
Can y’all help me with dis question
kykrilka [37]

Pretty sure is A or D.

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
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