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3 years ago

Please help me with these questions, time sensitive

Mathematics

1 answer:

Finger [1]3 years ago

3 0

Answer:

1. f(x)=−(x−3)(x+1)

By multiplying the factors, the general form is f(x)= -x²+2x+3.

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

So, Vertex = (1, 4)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

Now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

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3 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

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k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

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Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

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$Ae^{rt}(r^2-11r+30)=0$

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3 years ago

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