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ella [17]
3 years ago
14

I need help with this last question!!!!

Mathematics
2 answers:
patriot [66]3 years ago
7 0
••••••••••••••••••••••••••••••••••• it would be 43
Mariana [72]3 years ago
7 0

Answer:

h=43

Step-by-step explanation:

t=35

35/5 =7

50-7 =43

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Solve.<br><br> 5x-2y&lt;10<br> I give Brainliest!
Anestetic [448]
<h2>Answer:</h2>

This is impossible to solve.

<h2>Step-by-step explanation:</h2>

For an equation or inequality to be solvable, there must be the same number of inequalities as variables. Here, there is an x and there is a y. This means that you need at least two inequalities to solve it.

You can, however, rearrange to get x or y on one side.

This can be done for x:

5x < 10 + 2y

x < 2 + 2/5y

Or it can be done for y:

5x < 10 + 2y

5x - 10 < 2y

2.5x - 5 < y

8 0
2 years ago
HELP IM LITERALLY SOBBING
Vika [28.1K]

Answer: Y= 8x + 2.8

Step-by-step explanation: slope is 8 (think rise over run) and y-intercept seems to be 2.8, heh hope that helped

3 0
3 years ago
Read 2 more answers
in this diagram, AB and CD are Parallel. Angle ABC measure 35 and Angle BAC measure 115, what is the angle of ACE.
ZanzabumX [31]

Answer:

180°

Step-by-step explanation:

I think this is the answer

because the line is straight from the point ACE

see

5 0
3 years ago
Halp me again plez . . . . . . . . . . .
Katena32 [7]

Answer:

B?

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6 0
3 years ago
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The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

Substituting into the ODE, we get

v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0
v''x\ln x+(2+\ln x)v'=0

Setting w=v', we end up with the linear ODE

w'x\ln x+(2+\ln x)w=0

Multiplying both sides by \ln x, we have

w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x

we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
w=\dfrac{C_1}{x(\ln x)^2}

Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
v=-\dfrac{C_1}{\ln x}+C_2

So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
4 0
3 years ago
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