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3 years ago

The model represents an equation what value of X makes the equation true

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

A Solution is a value we can put in place of a variable (such as x) that makes the equation true.

Step-by-step explanation:

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When the expression 10^2021 − 2021 is evaluated, the result is a very large number.

Dafna11 [192]

Answer:

18,176

Step-by-step explanation:

Figure out an easier thing to do: 1000 - 2021 = 9,797.

Therefore, our answer will be like 9999999999999999999999997979. Out of all of these digits, there are only 2 that arent nines. We know the amount of digits since it's 10^2021 power. We will have one less since we are subtracting 2021. Therefore, 2020 digits in total. Now we know of those 2020 digits, 2 of them are 7, therefore 2018 of them are 9s. That means do 2018(9) + 2(7) = 14 + 18,162 = 18,176.

6 0

2 years ago

Write a compound inequality to represent all of the numbers between -4 and 6.

svetlana [45]

-4<x<6 would be the compound inequality

6 0

2 years ago

The first soccer team scored 8 goals in 4 games. The second soccer team scored 27 goals in 9 games. Which soccer team scored mor

Hatshy [7]

Answer:

The Second Soccer Team

Step-by-step explanation:

First Team: 8/4=2

Second Team: 27/9=3

The first team scored 2 goals per game while the second team scored 3 goals per game.

5 0

3 years ago

Read 2 more answers

2.4n + 1.2n - 0.12 = 7.08

Basile [38]

Answer:

n=2

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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

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