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3 years ago

15

(X-3)²=-49solve for x

1 answer:

rjkz [21]3 years ago

3 0

X=10 because. (X-3)=-49. so (10-3) with the power of 2=-49

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Help me find the arc length of this. Thank you.

BabaBlast [244]

The red chord is a diameter of the circle, so the length of the blue arc is half the circle's circumference.

The circumference of a circle with diameter $d$ is $\pi d$ , so the circle has circumference $8\pi$ , and so the arc has length $4\pi$ .

4 0

3 years ago

Mars2501 [29]

BOOKMARK is the answer to this question

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2 years ago

Triangle ABC has an Area of 4 3/8 square feet. If the length of AC IS 1 2/3 feet, what is the length of BC?

Kaylis [27]

Answer:

length of BC = 5 1/4 feet

Step-by-step explanation:

Area of a triangle = 1/2×base×height

Area = 4 3/8 square feet

Base = AC= 1 2/3 feet

Height = BC = ?

Area of a triangle = 1/2×base×height

4 3/8 = 1/2 × 1 2/3 × height

35/8 = 1/2 × 5/3 × height

35/8 = 5/6 × height

Height = 35/8 ÷ 5/6

= 35/8 × 6/5

= (35 × 6) / (8 × 5)

= 210 / 40

= 5 1/4 feet

Height = BC = 5 1/4 feet

length of BC = 5 1/4 feet

8 0

3 years ago

find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou

Neko [114]

Answer:

$v = \frac{32\pi}{3}$

or

$v=33.52$

Step-by-step explanation:

Given

$f(x) = 4x - x^2$

$g(x) = x^2$

$[a,b] = [0,2]$

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

Integrate

$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

Substitute values for a, b, f(x) and g(x)

$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

$v = 8\pi * \frac{4}{3}$

$v = \frac{32\pi}{3}$

or

$v=\frac{32}{3} * \frac{22}{7}$

$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

8 0

3 years ago

What is 4.2e+16 written out?

worty [1.4K]

42000000000000000

....................

4 0

3 years ago