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3 years ago

Simplify using the laws of exponents. [2⁴/2³]⁵

Mathematics

1 answer:

jonny [76]3 years ago

6 0

To know this answer we need to know how to solve it. Many people skip this part which makes their answer wrong.

It is asking us to simplify it using exponents.

Exponents are these little numbers beside a big number Telling it how much you multiply it by.

EXAMPLES

2^3 is just 2x2x2

2^4 is just 2x2x2x2

Notice the pattern? I hope so!

Now 2 to the power of 4 is 16. Our new equation looks like this

{16/2^3}^5.

Now 2 to the power of 3 is 8.

So, our new equation is

{16/8}^5

In which you get 2^5 which is just 2x2x2x2x2

And you get 32 as the final answer

I hope this helped and if you have questions just tell me!!

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Label the place value 7,854.209

yuradex [85]

7(thousands)
8(hundreds)
5(tens)
4(ones)
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2(tenths)
0(hundredths)
9(thousandths)
Hope this helps! I would appreciate it if I got brainliest! Thanks! =D

6 0

3 years ago

Read 2 more answers

Simplify the following expression.<br> 5 [13 + 10 = (3 + 2)] + 9 x 2

Sergio [31]

Answer:

i BELIEVE ITS

Step-by-step explanation:

5 [13 + 10 = (3 + 2)] + 9 x 2

65+50=25+18

115=43

7 0

3 years ago

A Simple Random Sample reflects that ________ individual in the population has an ________ chance of being selected.

monitta

Answer:

any, equal

Step-by-step explanation:

A Simple Random Sample reflects that any individual in the population has an equal chance of being selected.

7 0

3 years ago

In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion

lys-0071 [83]

$\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$

$\displaystyle
(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\
n=5\\
k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10$

----------------------------------------------------

$\displaystyle
(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\
n=5\\
k=4\\\\
\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90$

<u>It's 90</u>

6 0

3 years ago

PLEASE HELP ME!!!!!!!!!

vitfil [10]

$(x - 4) {}^{2} - 28 = 8 \\ (x - 4) {}^{2} = 8 + 28 \\ (x - 4) {}^{2} = 36$

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

$\sqrt{(x - 4) {}^{2} } = ± \sqrt{36} \\x - 4 = ±6 \\ x _{1}- 4 = 6 \: \: \: \: \: \: \: \: \: \: x _{2}- 4 = - 6 \\ x_{1} = 6 + 4 \: \: \: \: \: \: \: \: \: \: x_{2} = - 6 + 4 \\ x_{1} = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = - 2$

Well he started off correct to the point of completing the square.

$(x - 3) {}^{2} = 16 \\ x - 3 = ±4 \\ \: x_{1} - 3 = 4 \: \: \: \: \: \: \: \: \: \: \: \: x_{2} - 3 = - 4 \\ x_{1} = 7 \: \: \: \: \: \: \: \: \: \: x_{2} = - 1$

8 0

1 year ago