What is the slope-intercept equation of the line below ?

HELP!!!!!!!!!!! What are two possible measures of the angle below?

Andreyy89

Answer: First Option -90° and 630°

Solution:

1) In clockwise the angle is negative, then is -90°

2) Counterclockwise the angle is positive and it could be:

270°

270°+360°=630°

Then, the answer is -90° and 630°

6 0

3 years ago

Find the area of the parallelogram whose vertices are given below. A(negative 2,2) B(2,0) C(10,3) D(6,5)

swat32

Answer: Area of Parallelogram is 28 units squared.

Step-by-step explanation:

Knowing, any two vector sides of a parallelogram sharing the same initial point, we can find the area of parallelogram that two vectors are shaping, using the cross product of these two vectors a×b, where the area is given by

Area= |a×b| ___ (1)

From the given points,we may choose any three of them such and find a two vector and expresses the sides of parallelogram " or one side and a diagonal of parallelogram" ,we may choose for example B ,C and D.

Moreover, we may choose point B, to be common point for the two sides " the initial point of two vector" thus we need to find vector BD and vector BC .

Calculations are given in picture.

5 0

3 years ago

Can someone please do this for me?

katrin [286]

Answer:

x= 105

y= 40

Step-by-step explanation:

3y=4y-40 because they are alternate interior

Solve for y.

y=40

4y-40=x+15 because they are vertical angles.

Substitute 40 in for y.

4(40)-40=x+15

Solve.

x=105

5 0

2 years ago

Read 2 more answers

Find the values of m and b that make the following function differentiable.

dimaraw [331]

Now, bearing in mind that, for a graph, to be "differentiable", means a "smooth transition", not an abrupt edge or a "cusp", the first subfunction of the piece-wise is just a quadratic, the second subfunction is a linear, well, so the linear needs to hit the quadratic "smoothly", not make an abrupt edge, for that, the linear must continue the quadratic from 2 onwards, since 2 is the range edge.

now, what is the piece-wise when x = 2? well, f(2) = x² or (2)² or 4

that means, the linear needs to hit it at that point, to make it so, let's make it the slope of the linear, the same as the quadratic's.

what's the quadratic's slope? well, simple enough

$\bf \left. \cfrac{dy}{dx}=2x \right|_{x=2}\implies 4\impliedby \textit{this means }
\begin{array}{llll}
mx+b\\
\uparrow \\
4
\end{array}$

so we have y = 4x + b

now, recall, for the quadratic, when x =2, y = 4, so for the linear, when x = 2, y = 4 as well, this means

$\bf 4 = 4(2) + b\implies 4=8+b\implies 4-8=b\implies \boxed{-4=b}
\\\\\\
mx+b\implies 4x-4\\\\
-------------------------------\\\\
f(x)=
\begin{cases}
x^2&x\le 2\\
4x-4&x\ \textgreater \ 2
\end{cases}$

check the picture below.

4 0

2 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0

nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for x are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let a0 and a1 be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since a0=4a2 , I’ll choose a1 and a2 as the two arbitrary constants. We can still express all other constants in terms of a1 and/or a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all n≥0 , except for n=1 , since a1 is an arbitrary constant independent of a0 (and therefore independent of a2 ).

Plugging back into original power series for y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant a2 is =4+ a power series (starting at n=2 ). However, if we had the appropriate x -term, we would have a power series starting at n=0 . Since the other independent solution is simply y1=x, then we can let a1=c1−3c2, a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

brainly.com/question/11443401

#SPJ4

6 0

10 months ago