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3 years ago

write the formula of the parabola that has x-intercepts (1.3,0) and (-1.3,0), and y-intercept (0,169)

Mathematics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

y= 169 − $100^{2}$

Step-by-step explanation:

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2 years ago

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find the circumference of a circle that has an area of 452.16 square meters. (Use 3.1416 as the value of π.)

IceJOKER [234]

(1) circumference of a circle = 2π.R
(2) area of a circle = π.R²

Area = 452.16 m² (given), so we can calculate the radius:
452.16 = π.R²
452.16/π = R²

143.92 = R² → R =√143.92 ≈ 12 m
Having R = 12, now we can calculate the circumference :

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3 years ago

Find the perimeter

Inga [223]

Answer:

Step-by-step explanation:

Perimeter= 4S

Perimeter= 4(5)

Perimeter= 20

Perimeter 2= 2(L+W)

Perimeter 2= 2(11+7)

Perimeter 2= 22+14

Perimeter 2= 36

Total Perimeter= P1+P2

Total Perimeter= 20+36

Total Perimeter= 56

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3 years ago

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liberstina [14]

Answer:

x= +2, -2

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2 years ago

Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [

Darya [45]

Answer:

A=3 and B=6

Step-by-step explanation:

Increasing and Decreasing Intervals of Functions

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

$f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9$

Computing the first derivative:

$f' ( x ) = - 6 x^3 + 54 x - 108$

We find the critical points equating f'(x) to zero

$- 6 x^3 + 54 x - 108=0$

Simplifying by -6

$x^2 -9 x +18=0$

We get the critical points

$x=3,\ x=6$

They define the following intervals

$(-\infty,3),\ (3,6),\ (6,+\infty)$

Thus A=3 and B=6

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3 years ago