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SIZIF [17.4K]
3 years ago
11

Calculate the average mass of 100 marbles that weigh 824g

Mathematics
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

8.24 grams/marble

Step-by-step explanation:

Average mass = (total mass)/(number of marbles)

Here, that would be:

824 g

-------------------  =  8.24 grams/marble

100 marbles

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kifflom [539]
The answer is D. 62.5%

Hope this helped. Good luck!
6 0
3 years ago
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THIS IS WORTH 50 POINTS I will mark you as the brainlyest only if you graph this on a GRAPH. Graph this on a GRAPH y= 1/3x+3 .
jenyasd209 [6]
I hope this helps. Keep in mind the 1/3 is slope and the 3 is y-intercept.

3 0
3 years ago
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Need the answer asapp with the working out please thanks :)
patriot [66]
Hi,

The two numbers should be 12 and 30. 12=2x2x3 while 30=2x3x5.

Their HCF is 2x3=6 and their LCM is 2x3x2x5. Because of their HFC, we know that they are both multiple of 6. Also, the question says they both are GREATER than 6, so they can’t be 6 but are 6 times “something”. Thanks to the LCM, we know that “something” is equal to 2 for the first number and to 5 for the second one, the numbers hence being 12 and 30.

I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
7 0
3 years ago
Please help. no links
olga nikolaevna [1]

Answer:

144

Step-by-step explanation:

20×12=240

add 4 to 20 and 12

makes a new rectangle 24×16

24×16=384

384-240=144

7 0
3 years ago
If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
3 years ago
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