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alekssr [168]
3 years ago
12

Please help me out here *image attached

Mathematics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

Rate of change =  change y / change x

Change y = + 1

Change x = + 2

SO the rate of change is 1/2

WHAT IS THE Y VALUE WHEN X IS 9

y = 1 MULTIPLY by 4.5 = 4.5

x = 2 MULTIPLY by 4.5 = 9

Y = 4.5 when x = 9

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What is the solution to the system of equations
FromTheMoon [43]

Answer:

(0,3)

Step-by-step explanation:

solve for 2y

-X-2y= -6

2y=6+2x

substitute the given value of 2y into the equation -x-2y=-6

-x-(6+3x)= -6

solve for x

x=0

substitute the given value of X into the equation 2y=6+3x

2y=6+3×0

solve for y

y=3

the solution is (0,3)

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2 years ago
What is the relationship between the 4s in the number 4498
algol13
The 4 in the thousands place is 10 times the 4 in the hundreds place.
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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
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Answer:

\boxed{\bold{100 \ m^2}}

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Identify 'q'

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➤ \boxed{\bold{Mordancy}}

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You would take the amount of increase and divide it by the amount of total cost over the time period of certain months. Then you would multiply that number by the percentage of increase gained, and subtract 96.12 to get your answer.

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3 years ago
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