ANYONE PLEASE HELP ME I REALLY NEED THE ANSWER RIGHT NOW BECAUSE I HAVE TO PASS THIS LATER I HOPE Y'ALL CAN HELP ME:(

Mathematics

2 answers:

Zina [86]3 years ago

4 0

Step one
divide the coefficient
step two
use the quotient rule to simplify the exponents

d1i1m1o1n [39]3 years ago

3 0

Answer: When you divide powers that have the same base, you subtract the exponents.That's a pretty easy rule to remember! It's the opposite of the multiplication rule.When you multiply powers that have the same base, you add the exponents and when you divide powers that have the same base, you subtract the exponents.

Step-by-step explanation:

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What is the LCD of 5/6 and -3/8 ?

sergiy2304 [10]

5/6 and -3/8

Denominators: 6 and 8

We want a the smallest number 6 and 8 can get into.

6:

6, 12, 18, 24, 30, 36, 42, 48

8:

8, 16, 24, 32, 40, 48, 56, 64

As you can see, the smallest number that both 8 and 6 can get into is 24.

8 0

3 years ago

What is the value of 4 to the second power - 2 ( 3 times 5 + 1)

valentinak56 [21]

Answer:

4 to the second power = 16. that leaves us with 16 - 2 ( 3 x 5 + 1). next, multiply 3 x 5 = 15. that leaves us with 16 - 2 ( 15 + 1 ). next, add 15 + 1 = 16. that leaves us with 16 - 2 ( 16 ). next, multiply 2 ( 16 ) = 32. finally, subtract 16 - 32 = -16

Step-by-step explanation:

your answer is -16. hope this helps

8 0

3 years ago

Solve the following the inequality: 12 < 4x - 16 < 16

cluponka [151]

Answer:

7 < x < 8

Step-by-step explanation:

$12$

8 0

3 years ago

1/1×3 + 1/3×5 + ... + 1/47×49 HELP PLZ

Natalka [10]

Answer:

24/49

Step-by-step explanation:

Let's add the terms and see if there's a pattern

$\dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}=\dfrac{5+1}{1\times 3\times 5}=\dfrac{2}{5}\quad\text{sum of 2 terms}\\\\\dfrac{2}{5}+\dfrac{1}{5\times 7}=\dfrac{14+1}{5\times7}=\dfrac{3}{7}\quad\text{sum of 3 terms}$

Suppose we say the sum of n terms is (n/(2n+1)), the next term in the series will be 1/((2n+1)(2n+3)) and adding that to the presumed sum gives ...

$\dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n(2n+3)+1}{(2n+1)(2n+3)}=\dfrac{2n^2+3n+1}{(2n+1)(2n+3)}\\\\=\dfrac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\dfrac{n+1}{2n+3}\text{ matches }\dfrac{(n+1)}{2(n+1)+1}$