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2 years ago

Please y'all I need help

Mathematics

1 answer:

Andrej [43]2 years ago

7 0

Answer:

Step-by-step explanation:

I think I c cause one of the coordinates is in -4 and the other one is bw 10 and 8 :)

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How many 15° angles are in a circle? Enter your answer in the box. angles

Alex_Xolod [135]

Answer:

Step-by-step explanation:

a circle is 360

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360 divided by 15 =

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2 years ago

Add 49 to the quotient of 125 and 5

lakkis [162]

Answer is:

125/5+49

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3 years ago

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dalvyx [7]

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3 years ago

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Find the solution of the initial value problem dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

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3 years ago

Round 26.32 to the nearest tenths.

Snezhnost [94]

Answer:

26.3

It’s really simple actually

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3 years ago