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Basile [38]
3 years ago
7

How do i prove that this is a right triangle

Mathematics
1 answer:
Vera_Pavlovna [14]3 years ago
3 0

Answer:

B if its 90 degrees for point B.

Step-by-step explanation:

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I need help asap due tonight
riadik2000 [5.3K]

Answer:

4

Step-by-step explanation:

multiply 10 by w and 3

equation becomes 10w+30=70

subtract 30 by both sides

equation becomes 10w=40

divide both sides by 10

equation becomes w=4

your answer is 4!

7 0
3 years ago
Read 2 more answers
Tell whether the ordered pair is a solution of the linear system. (-4,-6); -3x+y=6, -2x+y=-8
harkovskaia [24]
Plug them in... -4=x -5=y

-3x-4+-6=6 (check)
-3x-4=12-6=6 (ordered pair works with this equation)

-2x-4+-6=-8 (check)
-2x-4=8-6=2 (ordered pair does not work with this equation)

Therefore the ordered pair IS NOT the solution.

Hope this helps :)

8 0
3 years ago
What is 5 - 3x + x - 32 in a simplified expression?
tresset_1 [31]

Answer:

-27-2x

Step-by-step explanation:

5-3x+x-32

(collect like terms)

5-32-3x+x

<h2><u><em>-27-2x</em></u></h2>
4 0
3 years ago
Read 2 more answers
What is $59.99 with 15% off
Serjik [45]
First you turn 15% into a decimal by dividing by 100
.15 then multiply by 59.99 and get around 8.99
Then subtract 8.99 from 59.99 and get $51
So $59.99 with 15% off is about $51

Hope this helped! :))
8 0
3 years ago
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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