15-9x in written expression

For each pair of points below, think about the line that joins them. For which pairs is the line parallel to the y-axis? Circle

gogolik [260]

Answer:

Option B.

Step-by-step explanation:

If two lines are parallel then their slopes are always same.

Following this rule we can find the slope by the given pairs of coordinates of the options.

If the slope of the line is same as the slope of y axis then the line passing through these points will be parallel to the y axis.

Slope of y - axis = ∞

Option A). Slope = $\frac{y-y'}{x-x'}$

= $\frac{24-8.5}{3.22-3.2}$

= $\frac{15.5}{0.02}$

= 775

Therefore, line passing through points (3.2, 8.5) and (3.22, 24) is not parallel to y axis.

Option B). Slope of the line passing through $(\frac{40}{3},\frac{14}{3})$ and $(\frac{40}{3},7)$ will be

= $\frac{\frac{14}{3}-7}{\frac{40}{3}-\frac{40}{3}}$

= ∞

Therefore, line passing though these points is parallel to the y axis.

Option C). Slope of the line passing through $(2.9,5.4)$ and (7.2, 5.4)

= $\frac{5.4-5.4}{7.2-2.9}$

= 0

Therefore, slope of this line is not equal to the slope of y axis.

Option B. is the answer.

5 0

4 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

3 years ago

Find value of y and x

Vika [28.1K]

Since the triangle is isosceles the measure of angle B is also 47.
since you have dropped a perpendicular bisector the measure of angle x is 90. and since the angles must add to 180 you get 47+47+a=180 and a= 86. since y is half this we have y is 43

if you like my answer please rate as brainliest! thx

4 0

3 years ago

Lisa is making her famous chocolate chip brownies. The recipe calls for 2 cups of chocolate chips, but Lisa can only find her La

Ilia_Sergeevich [38]

Answer:

I think the answer is 8

Step-by-step explanation:

7 0

3 years ago

You work at a bike shop. A customer purchases 6 new bike tires at $6.99 each. If the customer gives you $40, how much is still o

Mariulka [41]

Answer:

194 cents is the correct answer

Step-by-step explanation:

6.99 dollars is 41.94

7 0

3 years ago

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