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3 years ago

Identify the ordered pair that represents the vector from A(-8,-1) to B(-5,3) and the magnitude of vector AB.

1 answer:

wel3 years ago

8 0

In order to find the vector that points from A to B we need to subtract each component of A from the corresponding component of B, according to the formula: v(a→b)=(b1−a1,b2−a2) In this case we have : v(a→b)=(−5−(−8),3−(−1)) <span>v(a→b)=(3,4)

</span>To find the magnitude we use the formula: ||v|= √(v1^2)+(v1^2)

So:
||v|= √(32)+(42) ||v|= √9+16 ||v|= <span>√</span>25 ||v|= 5

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The graph shows the relationship between the pounds of candy bought at a candy shop and the total cost, in dollars.

Vesna [10]

C I took the test and got it right

6 0

2 years ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$

Solving for dy/dx:

$\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}$

Applying the first conditon (slope is zero)

$\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0$

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

$x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2$

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

$x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1$

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

4 0

3 years ago

Review the graph of function f(x).

s344n2d4d5 [400]

The two limits when x tends to zero are:

$\lim_{x \to \ 0^-} f(x) = 1\\\\ \lim_{x \to \ 0^+} f(x) = 0$

<h3 /><h3>How to get the limits when x tends to zero?</h3>

Notice that we have a jump at x = 0.

Then we can take two limits, one going from the negative side (where we will go along the blue line)

And other from the positive side (where we go along the orange line).

We will get:

$\lim_{x \to \ 0^-} f(x) = 1\\\\ \lim_{x \to \ 0^+} f(x) = 0$

Notice that the two limits are different, that means that the function is not a continuous function.

If you want to learn more about limits:

brainly.com/question/5313449

#SPJ1

6 0

2 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0

2 years ago

One bag contains three white marbles and five black marbles, and a second bag contains four white marbles and six black marbles.

Ket [755]

5/8 the first bag and 3/5 for the second bag.

7 0

3 years ago

Read 2 more answers