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borishaifa [10]
3 years ago
5

Find an equation in standard form for the ellipse with the vertical major axis of length 16 and minor axis of length 10.

Mathematics
2 answers:
Sedbober [7]3 years ago
7 0

Answer: The second answer

Step-by-step explanation:

Dimas [21]3 years ago
3 0

Answer:

<em>The first option. </em>

Step-by-step explanation:

You take the major vertical length and put it into this equation:

2a = 16

a/2 = 16/2

a = 8

You do the same with the minor length.

2b = 10

b/2 = 10/2

b = 5

You then switch the a^2 and the b^2 due to the fact that the veritcal length is longer than the horizontal. Then you use the formula to plug in 8 and 5.

(x - h)^2 / b^2 + (y - k)^2 / a^2 = 1

(x - h)^2 / 5^2 + (y - k)^2 / 8^2 = 1

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Can someone help me
erik [133]

Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

  (2x -12)/2 = arcsin(16/22×sin(24°))

  x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

  2x -12 = 0

  x -6 = 0 . . . . . divide by 2

  x = 6 . . . . . . . add 6

Then the range of values of x will be ...

  6 < x < 23.206

_____

* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

  16² = 2·AB²·(1 -cos(2x-12))

The value of AB drops out of the ratio of these equations, and the result for x is as above.

4 0
3 years ago
Read 2 more answers
Form a sequence that has two arithmetic means between -13 and 89. a. -13, 33, 43, 89 c. -13, 21, 55, 89 b. -18, -36, -72, -144 d
iogann1982 [59]
<span>Form a sequence that has two arithmetic means between -13 and 89. a. -13, 33, 43, 89 c. -13, 21, 55, 89 b. -18, -36, -72, -144 d. -18, -81, -144

Solution:

Since  it has to be between -13 and 89, letter d and b are not anymore considered to be the answer.

for a:
33-(-13)=46=d
43-33=10=d the value for this d is different from the two sequence,
89-43=46=d
they have different value for d, thus this is not the answer!

for c:
21-(-13)=34=d
55-21=34=d
89-55=34=d

they have the same value for d, thus the correct answer is </span><span> c. -13, 21, 55, 89</span>
8 0
4 years ago
What is the surface area of the prism?
Harman [31]
Area of lower rect + area of upper + 4 * sides rect areas

lower = upper rect

2 * rect1 + 4 * rect2

Area = 2 * 6 * 11 + 2 * 11 * 9 + 2 * 9 * 6 = 438 ft^2

or simply just take 2 ( and multiply each two sides.

5 0
3 years ago
Solve, using the substitution method.
ss7ja [257]

Answer:

(-6,-5)

Step-by-step explanation:

X+y=-11

solve for x

x=-11-y

substiute x value into 2nd equation

2x-3y=3

2(-11-y)-3y=3

-22-2y-3y=3

-5y=25

y=-5

substitute -5 for y in either equation to find x

x+y=-11

x+-5=-11

x=-6

8 0
3 years ago
Question 7
Eddi Din [679]

Answer:

Step-by-step explanation:

New height to initial height=

10/4=2.5

New width from initial width should be 5×2.5=12.5inches

4 0
3 years ago
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