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algol [13]
2 years ago
7

3/4 divided by 1/3 plz I need it

Mathematics
1 answer:
Annette [7]2 years ago
7 0

Answer:

0.25

Step-by-step explanation:

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Find the equation of the ellipse with the following properties. The ellipse with foci at (0, 6) and (0, -6); y-intercepts (0, 8)
sesenic [268]
Hi, thank you for posting your question here at Brainly.

The general equation of a horizontal ellipse is

(x-h)2/a2 + (y-k)2/b2 = 1, at center (h,k) while a = semi-major axis, b = semi-minor axis. These are related through the distance of the focus from the center,c. a2 = b2 + c2.

If you draw the points on a coordinate plane, the center of the ellipse is at (0,0), so h and k equals 0. Then, the minor axis (2b) spans from 8 to -8 of the y-axis. This is equal to 16 units. Hence,

2b = 16
b = 8
b^2 = 64

The distance between the two foci is 2c. Thus,

2c = 12
c = 6
c^2 = 36

Then, a2 = 64 + 36 = 100. Substituting to the general equation:

x^2/100 + y^2/64 = 1
8 0
3 years ago
Find the area of the rectangle.<br><br> l=10 in, w=7.5 in
Tpy6a [65]

Answer:

It should be 75 inches

Step-by-step explanation:

5 0
2 years ago
A bag contains 5 red marbles, 3 white marbles, and 4 blue marbles. what is the probability of randomly selecting a red marble, s
Dennis_Churaev [7]
Total number of marbles=5+3+4=12
Probability of picking a red marble =5/12
5 0
3 years ago
For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
The drama club has 120 members. 48
marissa [1.9K]
<h3>Answer:  40%</h3>

Work Shown:

48/120 = 0.40 = 40%

6 0
2 years ago
Read 2 more answers
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