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3 years ago

Which statement about g(x) = x^2 - 576 is true?

Mathematics

1 answer:

Zina [86]3 years ago

7 0

Answer:

The second option.

Step-by-step explanation:

Let's see if this will work.

(x+24)(x-24)

Let's open the brackets

x^2 - 24x +24x - 576

= x^2 - 576

Please give brainliest

I hope you understand. If you don't. Please, leave a comment below.

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What is 4 1/3 × 2 3/5? express it in lowest terms

soldier1979 [14.2K]

We have,

$4\dfrac{1}{3}\cdot2\dfrac{3}{5}=\dfrac{13}{3}\cdot\dfrac{13}{5}=\dfrac{13\cdot13}{3\cdot5}=\dfrac{169}{15}=\boxed{11.2\bar{6}}$

To express periodic number calculated we use:

$x=11.2\bar{6}\\10x = 112.\bar{6}\\100x=1126.\bar{6}\\90x=100x-10x\Longrightarrow90x=1126.\bar6-112.\bar6\\90x=1014\\x=\dfrac{1014}{90}=\boxed{11\dfrac{24}{90}}$

Hope this helps.

r3t40

3 0

3 years ago

Proving Pythagorean Theorem

MrRissso [65]

Answer:

Step-by-step explanation:

${a}^{2} + {b}^{2} = {c}^{2}$

ssubstitute a and b for any of the numbers (3 or 6) and get 36+9 which is 45

45 =

${c}^{2}$

so you would need to find the square root to get c

4 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

I have two bags of counters.

STatiana [176]

Answer:

a. 4/21

b. 4/7

Step-by-step explanation:

Please kindly check the attached files for explanation.

8 0

3 years ago

Read 2 more answers

Find the midpoint of the segment with the following endpoints. (2,9) and (8,1)

vazorg [7]

Answer:

$\displaystyle (5,5)$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Midpoint Formula: $\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Step-by-step explanation:

Step 1: Define

Point (2, 9)

Point (8, 1)

Step 2: Identify

(2, 9) → x₁ = 2, y₁ = 9

(8, 1) → x₂ = 8, y₂ = 1

Step 3: Find Midpoint

Simply plug in your coordinates into the midpoint formula to find midpoint

Substitute in points [Midpoint Formula]: $\displaystyle (\frac{2+8}{2},\frac{9+1}{2})$
[Fractions] Add: $\displaystyle (\frac{10}{2},\frac{10}{2})$
[Fractions] Divide: $\displaystyle (5,5)$

7 0

3 years ago