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3 years ago

Which function has the range (-infinity,-2]U[0,infinity)?

Mathematics

2 answers:

iren [92.7K]3 years ago

8 0

Answer:

B. y=csc(x)-1

Step-by-step explanation:

just got it right!

dolphi86 [110]3 years ago

6 0

Answer:

y= csc(x)+1

Step-by-step explanation:

A P E X

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What is an equation of the line that passes through the points (0, 3) and (5, -3)?

shepuryov [24]

Answer <u>(assuming it can be in slope-intercept form)</u>:

$y = -\frac{6}{5} x+3$

Step-by-step explanation:

1) First, use the slope formula $m =\frac{y_2-y_1}{x_2-x_1}$ to find the slope of the line. Substitute the x and y values of the given points into the formula and solve:

$m =\frac{(-3)-(3)}{(5)-(0)} \\m = \frac{-3-3}{5-0}\\m = \frac{-6}{5}$

So, the slope is $-\frac{6}{5}$ .

2) Now, use the slope-intercept formula $y = mx + b$ to write the equation of the line in slope-intercept form. All you need to do is substitute real values for the $m$ and $b$ in the formula.

Since $m$ represents the slope, substitute $-\frac{6}{5}$ for it. Since $b$ represents the y-intercept, substitute 3 for it. (Remember, the y-intercept is the point at which the line hits the y-axis. All points on the y-axis have an x-value of 0. Notice how the given point (0,3) has an x-value, too, so it must be the line's y-intercept.) This gives the following equation and answer:

$y = -\frac{6}{5} x+3$

3 0

2 years ago

Kim works 4 hours more each day than jill does, and jack works 2 hours less each day

Nadusha1986 [10]

Where's the rest of the question

6 0

3 years ago

Read 2 more answers

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ

liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$

We conclude that the line integral is 1/2

4 0

2 years ago

Find the missing length indicated?

kondor19780726 [428]

Answer:

D. 15

Step-by-step explanation:

Let the missing length be represented as x.

Thus:

(24 - x)/12 = x/20 => angle bisector theorem

Cross multiply

20(24 - x) = x(12)

480 - 20x = 12x

480 - 20x + 20x = 12x + 20x

480 = 32x

480/32 = 32x/32

15 = x

Missing length = x = 15

6 0

2 years ago

find x if the distance between points L and M is 15 and point M is located in the first quadrant. L=(-6,2) M=(x,2)

aivan3 [116]

Answer:

Therefore,

$x = 9$

Step-by-step explanation:

Given:

Let,

point L( x₁ , y₁) ≡ ( -6 , 2)

point M( x₂ , y₂ )≡ (x , 2)

l(AB) = 15 units (distance between points L and M)

To Find:

x = ?

Solution:

Distance formula between Two points is given as

$l(LM)^{2} = (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Substituting the values we get

$15^{2}=(x--6)^{2}+(2-2)^{2}\\\\225=(x+6)^{2}$

Square Rooting we get

$(x+6)=\pm \sqrt{225}=\pm 15\\\\x+6 = 15\ or\ x+6 = -15\\\\x= 9\ or\ x = -21$

As point M is located in the first quadrant

x coordinate and y coordinate are positive

So x = -21 DISCARDED

Hence,

$x = 9$

Therefore,

$x = 9$

3 0

3 years ago