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4 years ago

This says to multiply the special products how do I answer this problem? (x+8)(x-8)

Mathematics

1 answer:

meriva4 years ago

8 0

(x + 8)(x - 8)
x(x - 8) + 8(x - 8) =
x^2 - 8x + 8x - 64 =
x^2 - 64 <===

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Write an expression for “Triple the sum of 33 and b”. *

IrinaK [193]

Answer:

99+3b

Step-by-step explanation:

The original equation would be 3(33+b), but simplifying it would get 99+3b.

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3 years ago

there are 300 student and teachers at a sports event.Grade k to 2 is students: 137 Grade 3 to 5 students: 145 Teachers: ? what i

AVprozaik [17]

Answer:

Percentage of teachers = 6%

Step-by-step explanation:

Given;

Number of grade k-2 students = 137

Number of grade 3-5 students = 145

Total = 300

Number of teachers = 300 - (137+145) = 18

Percentage of teachers = (number of teachers÷total)× 100%

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3 years ago

Which point would not be a solution to the system of linear inequalities shown

Strike441 [17]

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3 years ago

Melanie is making a piece of jewelry that is in the shape of a right triangle. The two shorter sides of the piece of jewelry are

ollegr [7]

Using Pythagoras’ theorem a2 + b2 = c2 you can find out the third (longest) side. (6x6) + (8x8) = 100 (the square root of 100 is 10) so ur final sum is 6+8+10 = 24 mm

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2 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago