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3 years ago

Anyone in 10th grade please help me because I have to pass this later at 10 pm:(

Mathematics

1 answer:

Alenkinab [10]3 years ago

6 0

18) $19m^8n^8-\left(4m^5n\right)\left(3m^3n^7\right)$

$=19m^8n^8-4m^5n\cdot \:3m^3n^7$

$=19m^8n^8-12m^8n^8$

$=7m^8n^8$

19) $\left(-5cd\right)\left(-3c^4d\right)-\left(7c^2d^2\right)\left(2c^3\right)$

$=5cd\cdot \:3c^4d-7c^2d^2\cdot \:2c^3$

$=15c^5d^2-14c^5d^2$

$=c^5d^2$

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3 years ago

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3 years ago

Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.

serious [3.7K]

The greatest number of sandwiches that Olivia can make is 12.

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<u>Solution:</u>

Given that, Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.

We have to find what is the greatest number of sandwiches that Olivia can make? And how many slices of cheese will be on each sandwich?

Let the number of sandwiches be "n"

Then, number of slices of ham per sandwich $=\frac{48}{n}$

And, number of slices of cheese per sandwich $=\frac{36}{n}$

As slices can’t be fractions, "n" should be a factor of 48 and 36. So now let us find H.C.F of 48 and 36 as we want greatest number of sandwiches that Olivia can make.

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, and 36

The common factors are those which appear on both factors of 48 and 36 are 1, 2, 4, 6, and 12.

Out of these common factors, 12 is the greatest

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3 years ago

7/28 in simplest form

vampirchik [111]

$\frac { 7 }{ 28 }$

28 is 4 times 7, so we will rewrite it like that.

$28=4\cdot 7\\ \\ \frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 }$

7 is 7 times 7, so we will rewrite it like that.

$\frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 } =\frac { 1\cdot 7 }{ 4\cdot 7 } =\frac { 1 }{ 4 } \cdot \frac { 7 }{ 7 } =\frac { 1 }{ 4 } \cdot 1=\frac { 1 }{ 4 }$

So the simplest form is,

$\boxed { \frac { 1 }{ 4 } }$

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3 years ago

Read 2 more answers