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3 years ago

Find an equation for the line with slope m=6/5 and which goes through the point (15,8)

Mathematics

1 answer:

Charra [1.4K]3 years ago

8 0

Answer:

The answer is

Step-by-step explanation:

To find an equation of a line given a point and slope we use the formula

where

m is the slope

( x1 , y1) is the point

From the question the point is (15,8) and the slope is 6/5

Substitute the values into the above formula and solve

That's

We have the final answer as

Hope this helps you

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Very easy! 15 points!

Irina-Kira [14]

Answer:

taxes, borrow money, regulate commerce, uniform rule of naturalization, and regulate the value.

8 0

2 years ago

Keith exercises x hours a week. Troy exercises three hours less than Keith and Joe exercises two times more than Troy. Which exp

muminat

If you would like to know which expression represents the amount of time Joe exercises, you can calculate this using the following steps:

Keith ... x hours a week
Troy ... three hours less than Keith: x - 3
Joe ... two times more than Troy: 2 * (x - 3)

The correct result would be D) 2 * (x - 3).

7 0

2 years ago

Read 2 more answers

Correct answers only please!

MrRissso [65]

Answer:

23.6°

Step-by-step explanation:

In this question we have to use some trigonometry to work out angle XVW. Since we are working out angles all of the trigonometric functions will have to be to the ⁻¹. The first thing we need to identify is will formula will we use out of the following:

Sin⁻¹ = Opposite ÷ Hypotenuse

Tan⁻¹ = Opposite ÷ Adjacent

Cos⁻¹ = Adjacent ÷ Hypotenuse

-------------------------------------------------------------------------------------------------------

10 = Hypotenuse

4 = Opposite

-------------------------------------------------------------------------------------------------------

We know which formula to use because the length won't be in the triangle for example we will be using the Sin triangle because we don't have an adjacent. If we don't have an adjacent then the other formula's won't work.

Now we substitute in the values to find the value of angle XVW

Sin⁻¹ = Opposite ÷ Hypotenuse

Sin⁻¹ = 4 ÷ 10

The value of angle XVW is 23.57817848

6 0

2 years ago

Is 5x-7y=3 standard form?

barxatty [35]

Yes it is. Ax+By=C is standard form. :)

7 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago