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umka2103 [35]
3 years ago
15

PLEASE HELP SUPER EASY GRAPHING PROBLEM, WILL MARK U BRAINLIEST

Mathematics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

Step-by-step explanation:

x- int

1/3x - 2/3 = 0

1/3x = 2/3

3(1/3x = 2/3)

x = 2

(2, 0)

y- int

y = 1/3(0) - 2/3

y = 0 - 2/3

y = -2/3

(0, -2/3)

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What is the median of the data set? {50, 54, 62, 48, 49, 52}<br><br><br><br> plz help
Stella [2.4K]
Q1=49
Median=51
Q3=54

if i'm not wrong
3 0
3 years ago
Read 2 more answers
What is the solution to y + 7 = -9 A. y = -16 B. y = -2 C. y = 2 D. y = 16
Helen [10]

Answer:

A. y = -16

Step-by-step explanation:

Hello!

What we do to one side of the equation we have to do to the other.

y + 7 = -9

We have to get y by itself so we have to get rid of the seven by doing the opposite of what it says

The opposite of addition is subtraction so we subtract 7 from both sides

y + 7 - 7 = -9 - 7

Solve

y + 0 = -16

Simplify

y = -16

The answer is A. y = -16

Hope this helps!

4 0
3 years ago
For what value of x is f(x) = 10?<br> (-1,6)(2,8)(3, 10) (5, 16)
Alex_Xolod [135]

Answer:

(3,10)

Step-by-step explanation:

From f(x)=10 you can see that the value of y,which is equal to f(x), is 10.

Therefore, the only option you could pick is that where y=10,which is the case for option (3,10).

7 0
3 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Plz help me I really need it
IrinaVladis [17]

Answer:

i think [after solving] that it is option B and C

Step-by-step explanation:

4 0
3 years ago
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