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Gnesinka [82]
2 years ago
11

15-9x in written expression

Mathematics
1 answer:
IrinaK [193]2 years ago
7 0

Answer:

Step-by-step explanation:

there u go

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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
Read 2 more answers
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