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3 years ago

Write the equation of the ellipse in standard form

Mathematics

1 answer:

AVprozaik [17]3 years ago

8 0

The equation ellipse in standard form is: (x-h)2a2+(y-k)2b2=1

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The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

nignag [31]

Assuming that by "16t2" you mean $16t^2$ :

First of all, let's compute the position at the two required times:

$d(2)=16\cdot 2^2 = 16*4 = 64$

$d(5)=16\cdot 5^2 = 16*25 = 400$

So, the ball traveled a total distance of $\Delta d = 400-64 = 336$ , during the time $\Delta t = 5-2 = 3$ .

The average rate of change is the ratio between these two:

$\frac{\Delta d}{\Delta t} = \frac{336}{3} = 112$

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3 years ago

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Solve x2 + 8x + 22 = 0 by completing the square.

Liula [17]

Answer:

Step-by-step explanation:

the solution of a squared equation is

x = (-b ± sqrt(b² - 4ac)) / (2a)

in our case

a = 1

b = 8

c = 22

x = (-8 ± sqrt(64 - 88))/2 = (-8 ± sqrt(-24))/2 =

= (-8 ± sqrt(4×-6))/2 = (-8 ± 2×sqrt(-6))/2 =

= -4 ± sqrt(-6) = -4 ± i×sqrt(6)

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3 years ago

Find the area of each sector.

sergey [27]

Answer:

D. $\frac{525\pi}{8}$ m²

Step-by-step explanation:

Area of the sector = ½*r²∅

Where,

Radius (r) = 15 m

Angle in radians (∅) = $\frac{7\pi}{12}$

Plug in the values into the equation

Area of sector = $\frac{1}{2} * 15^2 * \frac{7\pi}{12}$

Area of sector = $\frac{1}{2} * 225 * \frac{7\pi}{12}$

Area of sector = $\frac{1*225*7\pi}{2*12}$

Area of sector = $\frac{1,575\pi}{24}$

Simplify

Area of sector = $\frac{525\pi}{8}$

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3 years ago

Find the value of 4th terms by using 3n-2

Serga [27]

10 is the answer I think

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3 years ago

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Based in the diagram which statement is true

Charra [1.4K]

Answer:

14. congruent means "THE SAME" so it will be "A" because there all the same letter.

15. D

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3 years ago