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3 years ago

Ben said the graph of the inequality

Mathematics

1 answer:

Ksivusya [100]3 years ago

7 0

Answer:

Ben is incorrect

The solutions are x>-3 and x < 3

Step-by-step explanation:

we have

$-x^{2} +9 > 0$

Multiply by -1 both sides

$x^{2} -9 < 0$

Adds 9 both sides

$x^{2} < 9$

The solutions are

$x < 3$

and

$-x < 3$ -----> Multiply by -1 both sides ----> $x > -3$

therefore

The solution is the interval (-3,3)

Ben is incorrect

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Answer:

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Step-by-step explanation:

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2 years ago

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Describe the process for<br> graphing the equation<br> y =-1/2x- 1

AysviL [449]

the formula is y=mx+b

m is the slope

b is the y-intercept form

so the -1 is the y-intercept form. so you need to start at that point on the y axis. then the slope which is -1/2 comes into play. in my school we use rise/run. so from the point you marked at -1 you will go down 1 and to the right 2 points then where you stop is your next point (mark it) keep doing that. also go up the other way from the -1 point. go up 1 and to the left 2. then connect all of your points with a line. I hope this helped.

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3 years ago

Why do whole numbers with an exponent get bigger while fractions with an exponent get smaller.

ohaa [14]

Raising a whole number to a power means:

   Take (a number of things) a number of times, and then possibly
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3 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

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

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

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

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

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

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

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

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3 years ago