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Ulleksa [173]
3 years ago
7

Biff earned $45 working at happy days drive-in. He spent 1/3 of the money on gas for his car and 1/5 of it on the flowers for hi

s girlfriend. How much money does he have left for the big date? I know what the answer is but I do not know how to show my work or explain it. Please show me how you got the answer thank you very much I will make the points on this problem go up!

Mathematics
1 answer:
bixtya [17]3 years ago
7 0
Hope this helps and hope its not too late

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Irina-Kira [14]

Answer:

we conclude that the intervals in which the function P(t) is decreasing are:

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Step-by-step explanation:

At x< 1, the function P(t) is decreasing

  • A function p is an increasing function on an open interval if f(y) > f(x) for any two input values x and y in the given interval where y>x
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From the figure, it is clear that the function seems to be increasing

from (1, 3) and then 4 to onwards.

But it is clear that the function seems to be decreasing from the x < 1 and from the interval (3, 4).

Therefore, we conclude that the intervals in which the function P(t) is decreasing are:

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3 years ago
Kala found an icicle 26 inches long. How long is it in feet?
yarga [219]
About 2 feet and 2 inches
5 0
3 years ago
I will give brainliest if you answer all correctly ​
miskamm [114]

Answer:

I think 13 is C

Step-by-step explanation:

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3 years ago
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
Jean listed the steps involved in the life cycle of a gymnosperm. 1 - The pollen gets stuck in a sticky liquid. 2 - Sperm from t
Andrews [41]

Answer:

b

Step-by-step explanation:

8 0
2 years ago
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