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Sliva [168]
3 years ago
5

Pls help i need this done today

Mathematics
2 answers:
Alika [10]3 years ago
7 0
The answer is C like 3
nalin [4]3 years ago
6 0
The answer is C. line 3
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The Graph of a function g is shown below. Find the following: g(10)
Yanka [14]

Answer: the graph is not shown, but I'll try to explain how hypothetically, you would be able to graph g(10). If you are given something like an image of the graph, find slope and y-intercept first. The y-intercept is the point (0, #), and is on the y-axis. After finding what g(x) = mx + b <- slope intercept form is, substitute 10 for x, then solve.

Step-by-step explanation:

3 0
3 years ago
Select all the correct statements.
lisabon 2012 [21]

Answer:

C,D,E,F

Step-by-step explanation:

A amplitude is 2

B period is 4pi

8 0
2 years ago
Read 2 more answers
Yo!!!NEED HELP!!!!WILL GIVE BRAIN!!!!!!!
yawa3891 [41]
First, we are going to find if the function is odd or even. Remember that we can determine if a function is odd of even from its graph by looking at its ends; if both ends go to the same the direction, the function is even. If both ends go to opposite directions, the function is odd. At both ends, the graph of our function go towards the same direction, minus infinity, so we can conclude that our function is even.

Next, we are going to find the possible degree of our function. Remember that the possible degree of a function is the number of x-intercepts.
We can infer from our graph that the function intercepts the x-axis at least 6 times.

We can conclude that the correct answer is: even degrees of 6 or greater.
5 0
3 years ago
Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
3 years ago
The slope of a line is 0 and it contains the point (3,1)
Sedaia [141]

Answer:

<h2>y = 1</h2>

Step-by-step explanation:

The point-slope form of an equation of a line:

y-y_1=m(x-x_1)

<em>m</em><em> - slope</em>

<em>(x₁, y₁)</em><em> - point on a line</em>

We have <em>m = 0, (3, 1) → x₁ = 3, y₁ = 1</em>.

Substitute:

y-1=0(x-3)

y-1=0      <em>add 1 to both sides</em>

y=1

Other method.

If the slope is equal 0, then it's a horizontal line.

The equation of a horizontal line: <em>y = a</em>.

We have the point on a line (3, 1) → x = 3, y = 1.

Therefore the equation is <em>y = 1</em>.

3 0
3 years ago
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