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astraxan [27]
3 years ago
10

Which of the following functions is graphed below?

Mathematics
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

the answer is c

Step-by-step explanation:

I took it already i think it's that one .

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En considérant que DE est
skad [1K]

Answer:

si

Step-by-step explanation:

5 0
3 years ago
If possible step by step explanation. thank you!
elixir [45]

Answer:

A

Step-by-step explanation:

\frac{(3x^{2} )^\frac{1}{2} }{3^\frac{1}{2} } =3

(3x^2)^\frac{1}{2} = 3^\frac{1}{2} x

Remember that \frac{x^b}{x^c} =x^(b-c)

Using that \frac{3^\frac{1}{2}x }{3^\frac{1}{2} }=3^(1/2x-1/2)=3^((x-1)/2)

3^\frac{x-1}{2} =3^1

So we can say: \frac{x-1}{2} =1, because the bases are the same

We can multiply both sides of the equation by 2. We get x-1=2, and x=3. Which is A.

7 0
3 years ago
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
3 years ago
Find an equation of the tangent line to the hyperbola: x^2/a^2 - y^2/b^2 = 1 at the point (x0,x1)
musickatia [10]

The standard form of a hyperbola is <span><span><span>x2/</span><span>a2 </span></span>− y<span><span>2/ </span><span>b2 = 1
the tangent line is the first derivative of the function</span></span></span><span>y′ = <span>b^2x/ a^2 y
hence the slope is </span></span><span>m = <span>b^2 x0 / <span>a^2 <span>x1

</span></span></span></span>Therefore the equation of the tangent line isy−x1 = b^2 x0 / a^2 x1* (x−x0)

8 0
3 years ago
In Exercises 1-16, solve the inequality. Graph the solution<br> 1. 3y ≤ -9
murzikaleks [220]

By solving the inequality we get y\leq -3

What is graphing the inequality?

The act of illustrating which area of the number line contains values that will "satisfy" the specified inequality is known as graphing the inequality. Take a look at the first inequality, "x > -5." The numbers that can be used to substitute x in our inequality to produce a true statement can be shown on a graph of our inequality.

3y\leq -9

Dividing above equation by 3 we get,

\frac{3y}{3}\leq \frac{-9}{3}\\y\leq -3

Graph of y\leq -3 is

Therefore by solving the inequality we get y\leq -3

To learn more about graphing the inequality from the given link

brainly.com/question/24372553

#SPJ1

8 0
1 year ago
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