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klasskru [66]
3 years ago
14

Does the following equation have one ,no or infinite solution? 5(x+6) =5x+30

Mathematics
1 answer:
Mumz [18]3 years ago
5 0
Yes it does, let me try to answer it!
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36 increased by twice x is
ICE Princess25 [194]
36 + x  would be your answer. When something is 'increased', you are adding to it. Such as saying 47 increased by -5x is the same as 47 + (-5x).
If this wasn't the answer you were looking for, let me know. For now, I hope this helped:)

4 0
3 years ago
Read 2 more answers
Geometry math question no Guessing and Please show work
MAXImum [283]

If the point P(x, y) partitions the segment AB in the ratio 1 : 1, then the point P is midpoint of segment AB.

The formula of a midpoint of segment:

A(x_A;\ y_A),\ B(x_B,\ y_B)\\\\P_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)

We have:

A(13,\ 1)\to x_A=13,\ y_A=1\\B(-5,\ -3)\to x_B=-5,\ y_B-3\\\\P\left(\dfrac{13+(-5)}{2};\ \dfrac{1+(-3)}{2}\right)\to P\left(\dfrac{8}{2};\ \dfrac{-2}{2}\right)\to P(4,\ -1)

Answer: C. (4, -1)

3 0
3 years ago
How many solutions does the equation 9x-3(x+8)=6x-24 have?
Neporo4naja [7]
I think the answer is C-No solutions
4 0
3 years ago
What is 40,023,032 in expanded form
Tanya [424]

40,023,032 = (4 x 1000000000) + (0 x 100000000) + (0 x 10000000) + (0 x 1000000) + (2 x 100000) + (3 x 10000) + (0 x 1000) + (0 x 100) + (3 x 10) + (2 x 1)

4 0
3 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t > 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
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