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3 years ago

15

Draw a line with a negative slope

2 answers:

lubasha [3.4K]3 years ago

7 0

Y=-x+2 is an equation that you can use

Marina86 [1]3 years ago

4 0

Sorry, I didn't fully understand your initial question. I'll answer it now.

A quick and easy equation you can always use is y = -x + 2

Put one point on (0, 2) and put another point on (2, 0) (since a slope of 2/2 would cancel out) Then connect the points, and you're done.

You might be interested in

List the perfect squares that are between 25 and 100.

Aleks04 [339]

Perfect squares are rational numbers multiplied by themselves. Here are the perfect squares between 25 and 100:

5*5=25

6*6=36

7*7=49

8*8=64

9*9=81

10*10=100

Hope this helps :)

7 0

3 years ago

Read 2 more answers

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Find the value of x

Alexeev081 [22]

Answer:

75

Step-by-step explanation:

180-30=150

150/2=75

6 0

2 years ago

Select the equation that contains the point (-5, 2) and in which the slope equals 1/2

goblinko [34]

C is the correct answer. using the equation (y-y1)=m(x-x1)

6 0

2 years ago

WILL MARK BRAINLIEST, JUST HELP ME PLS

Luda [366]

Answer:

SAS theorem

Step-by-step explanation:

If $\frac{CE}{CB} = \frac{DC}{AC}$ and m∠ECD = m∠BCA, then the triangles are similar by the SAS theorem.

$\frac{EC}{BC} = \frac{14}{21} = \frac{2}{3}$ $\frac{DC}{AC} = \frac{18}{27} = \frac{2}{3}$

So, $\frac{CE}{CB} = \frac{DC}{AC}$ and m∠ECD = m∠BCA because vertical angles are congruent

Therefore ΔECD is similar to ΔBCA by SAS theorem

5 0

2 years ago