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timofeeve [1]
2 years ago
12

Please, help me ,,,,,,,,,,,,,,,

Mathematics
2 answers:
Olin [163]2 years ago
8 0
I will probably be the last one but I am not sure
LuckyWell [14K]2 years ago
3 0

Answer:

The correct answer is 80, 190

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The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
3 years ago
Please write eight facts about: The Flow of Energy in a Food Chain.
Elan Coil [88]

Answer:

Always linear...

Decreases with successive trophic level

5 0
2 years ago
Read 2 more answers
EXPRESS in power notation
lora16 [44]

Answer:

Solution given:

1:

-1/8=\frac{-1³}{2³}=(\frac{-1³}{2³}=-1³*2^{-3}=-2^{-3}

2. -64/27

=\frac{-4^{3}}{3³}=\frac{-4³}{3³}=(\frac{-4}{3})^{3}

Express in rational number

1. (-3/2)=-3/2*2/2=-(3*2)*(2/2)=-6/4

2. (1/5)=1/5*5/5=<u>5/25</u>

and

(-4/3)³(2/5)-⁴ ÷ (7/4)

(-4³/3³)(2-⁴/5-⁴)÷7/4

(-64/27)(5⁴/2⁴)÷7/4

(-64/27)(625/16)÷7/4

(-64*625/(27*16))*4/7

-2500/27*4/7

-10000/169

(-100/13)²

7 0
2 years ago
*** WHITE Given m|n, find the value of x. (2x-3) (6X+17)​
-BARSIC- [3]

Answer:I dont belive i know

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A ball is thrown vertically upward from the top of a building 160 feet tall with an initial velocity of 48 feet per second. The
alina1380 [7]

Answer:

# after 5 seconds, the ball strikes the ground

# The ball reaches maximum height at t = 1.5 seconds

# The max height is 196 feet

Step-by-step explanation:

The equation is:

d(t)=-16t^2+48t+160

t is the time

d(t) is the distance traveled

Initial height is 160 feet and initial velocity is 48 ft/sec

If we want to find the time it takes the ball to hit the ground, we let d(t) equal to 0 and find t. Shown below:

-16t^2+48t+160=0\\t^2-3t-10=0\\(t-5)(t+2)=0\\t=5,-2

We disregard t = -2 since time can't be negative. We take t = 5

Thus, after 5 seconds, the ball strikes the ground.

The equation is a quadratic of the form: ax^2+bx+c

Matching equations, we can say:

a = -16

b = 48

c = 160

The time when ball reaches max height is given as:

t=-\frac{b}{2a}

Substituting, we find:

t=-\frac{b}{2a}\\t=-\frac{48}{2(-16)}\\t=1.5

The ball reaches maximum height at t = 1.5 seconds

The max height can be found by putting t = 1.5 into the original equation. Shown below:

d(t)=-16t^2+48t+160\\d(1.5)=-16(1.5)^2+48(1.5)+160\\=196

The max height is 196 feet

8 0
3 years ago
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