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3 years ago

Express 12113 in base 10

Mathematics

1 answer:

dlinn [17]3 years ago

6 0

Answer:

log 12113 in base 10

hope this will help you

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Find the result of 765 times 95 divided by 5? please help me

Marizza181 [45]

Answer:

14,535

Step-by-step explanation:

765 times 95 : 765 x 95 = 72,675

divided by 5 : $\frac{ 72,675}{5} = 14,535$

Let me know if you want an even more in-depth answer

Happy to do so !

7 0

2 years ago

I do not understand this question, anyone help? Most brainiest for the right answer!

Oksi-84 [34.3K]

Answer:

56.7

Step-by-step explanation:

We know

mean = sum / count, with count being the amount of papers corrected in this case. We want to find the sum of all the papers as well as the count to figure out the mean of all papers.

For Tony's papers,

mean = sum / count

50 = sum / 40

multiply both sides by 40 to isolate the sum

sum = 50 * 40 = 2000

For Alice's papers,

mean = sum / count

70 = sum / 20

multiply both sides by 20 to isolate the sum

70 * 20 = sum = 1400

The total sum of all 60 papers is equal to the sum of 40 papers + the sum of the remaining 20 papers, or 2000 + 1400 = 3400. The mean of the 60 papers is therefore

mean = sum / count

mean = 3400/60 ≈ 56.7

6 0

3 years ago

I need to know what m n and q are equal to

VMariaS [17]

I know m=2.25 and n=8.6 but I don't know about q.

3 0

3 years ago

If a certain number of Grade IV students share 28 or 39 skill books in Mathematics,there are remainders of 4 and 3 books. What i

nadya68 [22]

Answer:

12 students

Step-by-step explanation:

Below numbers are divisible by the number of students:

28 - 4 = 24
39 - 3 = 36

Lets find GCF of 24 and 36:

24 = 2*2*2*3
36 = 2*2*3*3
GCF(24,36) = 2*2*3 = 12

The largest possible number of students is 12

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2 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

Step-by-step explanation:

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

2 years ago

Express 12113 in base 10​

Express 12113 in base 10