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4 years ago

What is the y-intercept of the graph below? A. (0, 4) B. (0, -2) C. (0,2) D. (0,4)

Mathematics

2 answers:

atroni [7]4 years ago

6 0

Answer:

Step-by-step explanation:

The y- intercept is the point where the graph crosses the y- axis.

The graph crosses the y- axis at (0, 2 ) → C

irga5000 [103]4 years ago

3 0

Answer:

C. (0,2)

the line goes through y 2 at x 0

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Help please this shouldn't take long to answer I'm just bad at this this stuff (brainliest)

icang [17]

Answer:

5 - 17 - 25 - 42 - 64

Step-by-step explanation:

<u>Given data set: </u>

5, 14, 20, 23, 25, 29, 40, 44, 64

It is already in ascending order.

<u>As per data the 5-number summary is:</u>

Minimum = 5
First quartile = (14 + 20)/2 = 17, the median of the lower half
Median = 25, the middle number as the set comprises of 9 numbers
Third quartile = (40 + 44)/2 = 42, the median of the upper half
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3 years ago

A bell tolls every 10 minutes. Another bell tolls every 15 minutes. Both bells toll at 6:00 PM. They will toll together again at

Verizon [17]

Im pretty sure they will both ring at the same time at 6:30

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3 years ago

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Change your money from dollars to british pounds.you got 560 pounds for 800 dollars.how many pounds will you get for 300 dollars

Anettt [7]

You do a proportion so 560x300=168,000 so then you take 168,000 and divide by 800 so the answer is 210 you would have 210 British pounds for 300 dollars

7 0

3 years ago

What is the y-intercept of the line given by the equation y = 9/2x -4 Enter your answer as an ordered pair.

V125BC [204]

Answer:

(0, -4)

Step-by-step explanation:

I don't quite know how to explain it, but I was taught that the number in the equation that doesn't have a variable, in this case -4, is the y-intercept. The order pair is (0, -4) because a y-intercept is where the line crosses the y-axis and the y-axis is located on the 0 of the x-axis.

6 0

3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago