3 years ago

Please help asap thanks in advance.

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

hope this helps

please like and Mark as brainliest

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Write in expanded form 51.242.

lord [1]

Answer:

50 + 1 + $\frac{2}{10}$ + $\frac{4}{100}$ + $\frac{2}{1,000}$

Step-by-step explanation:

This is the answer because:

1) 50 and 1 equal to 51

2) The fraction forms of a decimal number is: the decimal number/the decimal number place value

3) Therefore, the answer is 50 + 1 + $\frac{2}{10}$ + $\frac{4}{100}$ + $\frac{2}{1,000}$

Hope this helps!

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3 years ago

A company requires a password to access a data storage vault. The password must be 4 characters long and may

lisov135 [29]

Answer:

There are 36 possibilities per character (26 lower case letters + 10 numbers), so the total number of permutations are 36^4 = 1,679,616

Step-by-step explanation:

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3 years ago

DO NOT REDUCE Your Answer To The Lowest Terms! What Is 3/4 x 2 and 3/4

natka813 [3]

33/16 I’m sorry if it’s wrong

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3 years ago

Solve asap: 3(2y+10)=54 Within 10 mins pls 8th grade

inna [77]

Answer:

y = 4

Step-by-step explanation:

6y + 30 = 54

subtract 30 from each side

6y = 24

divide each side by 6

y = 4

7 0

3 years ago

Read 2 more answers

I'll mark brainliest just tell me how to.

IRINA_888 [86]

Answer:

a = 12

b = 2

c = 11

Step-by-step explanation:

$\frac{( {x}^{5}y {z}^{4} )^{3} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ \frac{ {x}^{5 \times 3}y^{3} {z}^{4 \times 3} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ \frac{ {x}^{15}y^{3} {z}^{12} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ {x}^{15 - 3} {y}^{3 - 1} {z}^{12 - 1} = {x}^{a} {y}^{b} {z}^{c} \\ \\ {x}^{12} {y}^{2} {z}^{11} = {x}^{a} {y}^{b} {z}^{c} \\ \\ equating \: like \: terms \: from \: both \: sides \\ \\ {x}^{12} = {x}^{a} \: \implies \: a = \boxed{ 12}\\ \\ {y}^{2} = {y}^{b} \: \implies \: b = \boxed{ 2} \\ \\ {z}^{11} = {z}^{c} \: \implies \: c = \boxed{ 11}$

3 0

3 years ago

Please help asap thanks in advance. ​

Please help asap thanks in advance.