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Sidana [21]
4 years ago
12

Write an equation in point slope formula of the line that passes through point (-3,0) and has a slope of -1/3

Mathematics
1 answer:
olasank [31]4 years ago
8 0

Answer:

Step-by-step explanation:jdjdjsjsjsjssjjsns

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How to write 8 17/20 as a decimal number
worty [1.4K]

Hello!

Answer:

In order to convert mixed number 8 17/20 to decimal form, you follow these steps:

Convert the proper fraction to a decimal: Divide the numerator by the denominator

17 ÷ 20 = 0.85

Add this decimal number to the whole number of the mixed fraction

8 + 0.85 = 8.85

The mixed number 8 17/20 converted to decimal number is, therefore:

8.85

Hope this helps <3

4 0
3 years ago
Read 2 more answers
Prove that the point [1,2] is equidistance from the point [4,-2], [5,-1] and [-2,6] ​
fgiga [73]

Step-by-step explanation:

Proof: equidistance means at an equal distance from two points

so: Take the distance formula and then plug the values in and compare.

3 0
2 years ago
SENPAI ANSWER THIS TOO. 2X+2X=?
natulia [17]

Answer: 4x

2x + 2x=4x

7 0
3 years ago
Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 7 liters of natural oil for every 5 liters of
Svetach [21]
What we know:
Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 7 liters of natural oil for every 5 liters of synthetic oil.

What we need to find:
If 355 liters of synthetic oil are used, how many liters of petrolyn oil will be made?

Set up a proportion to find how much natural oil is used.
7L natural oil/5L synthetic oil = x/355 synthetic oil where x=natural oil
7/5=x/355
(5)(x)=(7)(355) cross multiplied
5x=2485 simplified
5/5x=2485/5 multiplicative inverse
x=497L

Now, add both synthetic oil and natural oil to find amount of petrolyn oil.

355 L synthetic oil + 497 L natural oil= 852 L of petrolyn oil
5 0
3 years ago
Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)
rosijanka [135]

We can try reduction order and look for a solution y_2(x)=y_1(x)v(x). Then

y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v

Substituting these into the ODE gives

x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0

x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0

x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0

which leaves us with an ODE linear in w(x)=v'(x):

x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0

This ODE is separable; divide both sides by the coefficient of w'(x) and separate the variables to get

w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0

\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}

\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx

Integrate both sides; on the right, substitute u=\ln x so that \mathrm du=\dfrac{\mathrm dx}x.

\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du

Now solve for w(u),

\ln|w|=-2\ln(\cos u)-u+C

w=e^{-2\ln(\cos u)-u+C}

w=Ce^{-u}\sec^2u

then for w(x),

w=Ce^{-\ln x}\sec^2(-\ln x)

w=C\dfrac{\sec^2(\ln x)}x

Solve for v(x) by integrating both sides.

v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx

Substitute u=\ln x again and solve for v(u):

v=\displaystyle C_1\int\sec^2u\,\mathrm du

v=C_1\tan u+C_2

then for v(x),

v=C_1\tan(\ln x)+C_2

So the second solution would be

y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)

y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)

y_1(x) already accounts for the second term of the solution above, so we end up with

\boxed{y_2=x^2\sin(\ln x)}

as the second independent solution.

6 0
4 years ago
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