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Kazeer [188]
3 years ago
10

Factor: 24x^3−81 x^3 is x to the third power

Mathematics
1 answer:
MArishka [77]3 years ago
8 0
Hi again!

24x³ - 81

= 3(2x -3)(4x²+6x+9)


I hope that helps!

Please Brainliest answer :)
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How are the values of the eights in 880 related
Neko [114]
<span>How the value of 8 in 880 related
=> 880 = 8 hundreds 8 tens
=> This given number is a whole numbers, thus the value of 8 in the hundreds place is ten times larger than the 8 In the tens place. How?
Try multiplying 8 tens by 10
=> 8 tens = 80
=> 80 x 10
=> 800
When 80 is multiplied by ten the answer is 800 and when 800 is divided by 10 the answer is 80.
That concludes that there’s 10 times difference between the two 8s in the given number.</span><span>


</span>



8 0
3 years ago
What is the value of the expression 50 - (5 + 2)2 + 6?
umka2103 [35]

Answer:

I believe it's 42

Step-by-step explanation:

Replace each letter in the expression with the assigned value.

First, replace each letter in the expression with the value that has been assigned to it. To make your calculations clear and avoid mistakes, always enclose the numbers you're substituting inside parentheses. The value that's given to a variable stays the same throughout the entire problem, even if the letter occurs more than once in the expression.

However, since variables "vary", the value assigned to a particular variable can change from problem to problem, just not within a single problem.

Perform the operations in the expression using the correct order of operations.

Once you've substituted the value for the letter, do the operations to find the value of the expression.

7 0
2 years ago
A piece of fabric cost $529 a yard which is an estimate cost of 16.3 yards of fabric ?
Drupady [299]

$8622.70 or $10,000 if you are actually estimating.

5 0
3 years ago
The difference between thirty and five
Rudiy27
The difference bettween 30 and 5 would be 25
5 0
3 years ago
Read 2 more answers
Use induction to prove: For every integer n &gt; 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

3 0
3 years ago
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