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Julli [10]
4 years ago
13

Is this a function or no

Mathematics
2 answers:
aev [14]4 years ago
7 0

Answer: Yes it is a function

Step-by-step explanation:

stellarik [79]4 years ago
6 0

Answer:

YES

Step-by-step explanation:

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Please help <br><br> Please <br><br> Which 2 segment are parallel? <br><br> Please
Alla [95]

Answer:

E and D

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Three students, Henry, Sarah, and Kylie, line up one behind the other. How many
mrs_skeptik [129]

Answer:

There are 6  different ways

Henry, Sarah, and Kyle

Henry, Kyle, and Sarah

Sarah, Henry, and Kyle

Sarah, Kyle and Henry

Kyle, Henry, and Sarah

Kyle, Sarah. and Henry

Step-by-step explanation:

Those are the orders they can line up in, There are I think 6 different ways they can stand in a line. Hope this helps, please tell me if I'm wrong in the comments, since it well help me on improving in my education.

8 0
3 years ago
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
4 0
3 years ago
David keeps records of his monthly earnings. Last month he worked 120 hours and earned $2700. This month he worked 100 hours and
lianna [129]
F(x) = 20x+300
you sub in the x values (120 hours and 100 hours) for each equation
the only one they both work for is the last one
f(x) = 20x + 300
f(100) = (20 * 100) + 300 = 2000 + 300 = 2300

f(120) = (20 * 120) + 300 = 2400 + 300 = 2700
8 0
3 years ago
Read 2 more answers
Raymond normally has a snack consisting of ten 12-calorie crackers. How many 20-calorie cookies would he have to eat to consume
ss7ja [257]
<h3>Answer:  6</h3>

Work Shown:

He eats ten 12-calorie crackers, so that is 10*12 = 120 calories total

If x is the number of 20-calorie cookies, then the total number of calories here is 20x. We want this to be 120, so,

20x = 120

x = 120/20 .... divide both sides by 20

x = 6

He must eat six 20-calorie cookies to get 120 calories total

3 0
4 years ago
Read 2 more answers
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