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2 years ago

Name the property the equation illustrates.

Mathematics

2 answers:

Mamont248 [21]2 years ago

7 0

Answer:

commutative property of addition

True [87]2 years ago

6 0

The answer is commutative property of addition

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Jose traveled 315 miles in 7 hours. Based on this rate, how many miles did Jose travel in one hour? Do not include units.

DiKsa [7]

Answer:

State your explanation below:

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2 years ago

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Given the second order homogeneous constant coefficient equation y′′−4y′−12y=0 1) the characteristic polynomial ar2+br+c is r^2-

Vitek1552 [10]

Answer:

The initial value problem $y(x) = 4 e^{-2x} -3 e^{6 x}$

Step-by-step explanation:

Step1:-

a) Given second order homogenous constant co-efficient equation

$y^{ll} - 4y^{l}-12y=0$

Given equation in the operator form is $(D^{2} -4D-12)y=0$

Step 2:-

b) Let f(D) = $(D^{2} -4D-12)y$

Then the auxiliary equation is $(m^{2} -4m-12)=0$

Find the factors of the auxiliary equation is

$m^{2} -6m+2m-12=0$

m(m-6) + 2(m-6) =0

m+2 =0 and m-6=0

m=-2 and m=6

The roots are real and different

The general solution $y = c_{1} e^{-m_{1} x} + c_{2} e^{m_{2} x}$

the roots are $m_{1} = -2 and m_{2} = 6$

The general solution of given differential equation is

$y = c_{1} e^{-2x} + c_{2} e^{6 x}$

Step 3:-

C) Given initial conditions are y(0) =1 and y1 (0) =-26

The general solution of given differential equation is

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$ .....(1)

substitute x =0 and y(0) =1

$y(0) = c_{1} e^{0} + c_{2} e^{0}$

$1 = c_{1} + c_{2}$ .........(2)

Differentiating equation (1) with respective to 'x'

$y^l(x) = -2c_{1} e^{-2x} + 6c_{2} e^{6 x}$

substitute x= o and y1 (0) =-26

$-26 = -2c_{1} e^{0} + 6c_{2} e^{0}$

$-2c_{1} + 6c_{2} = -26$ .............(3)

solving (2) and (3) by using substitution method

$substitute c_{2} =1- c_{1}$ in equation (3)

$-2c_{1} + 6(1-c_{1}) = -26$

on simplification , we get

$-2c_{1} + 6(1)-6c_{1}) = -26$

$-8c_{1} = -32$

dividing by'8' we get $c_{1} =4$

substitute $c_{1} =4$ in equation $1 = c_{1} + c_{2}$

so $c_{2} = 1-4 = -3$

now substitute $c_{1} =4 and c_{2} =-3$ in general solution

$y(x) = c_{1} e^{-2x} + c_{2} e^{6 x}$

$y(x) = 4 e^{-2x} -3 e^{6 x}$

now the initial value problem

$y(x) = 4 e^{-2x} -3 e^{6 x}$

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3 years ago

Find the range of the function f(x)=(x-2)^2+2

kari74 [83]

The function is in vertex form of a quadratic function.
Vertex form is f(x)=a(x-h)^2+k
The vertex is (h, k)

In this case, our vertex is (2, 2). The a value is positive which means that the graph points up. Since the graph points up, the vertex is the minimum, the lowest point of the graph. The range refers to y-values, so the range of the function is y/y>2. It can also be written as 2≤y≤positive infinity.

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2 years ago

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A) how many 15-year old students took part in the survey?

trapecia [35]

Answer: 60 or something near that sorry if wrong

Step-by-step explanation:

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2 years ago

A man spend 2500 per month for four months and runs into debts. He then reduces his expenses 1750 per month and just clears of h

nikdorinn [45]

Answer:

Step-by-step explanation:

Let the income per month be S

Total income in 8 months = 8 S .

Total expenses in four months = 2500 x 4 = 10000

Total expenses in next four months = 1750 x 4 = 7000

Total expenses = 17000

According to question

Total income in 8 months = Total expenses

8 S = 17000

S = 2125 per month .

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2 years ago