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2 years ago

PLEASE HELP ME I NEED HELP PLEASE HELP

Mathematics

2 answers:

lyudmila [28]2 years ago

6 0

Answer:

1st question is 1:2

2nd question is 5

Step-by-step explanation:

15 times two is 30, so if you divide both 15 and 30 by 15, you get 1:2

Divide 300 and 60 and you get 5.

strojnjashka [21]2 years ago

6 0

Answer:

1/2 5

Step-by-step explanation:

so the push ups you would divide 300 by 60 and the ratio is 1/2 because 15 plus 15 equals 30 so yea its 1/2

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creativ13 [48]

Using a system of equations, it is found that the unit prices are:

$2.25 for a bag of chips.

$1.50 for a liter of pop.

$1.75 for a chocolate bar.

For the system:

x is the unit price of a bag of chips.

y is the unit price of liter of pop.

z is the unit price for a chocolate bar.

From the table, the equations are:

$2x + 2y + z = 9.25$

$3x + 4y + z = 14.50$

$x + 3y + 3z = 12$

Replacing the first equation on the second and the third:

$z = 9.25 - 2x - 2y$

$3x + 4y + z = 14.50$

$3x + 4y + 9.25 - 2x - 2y = 14.50$

$x + 2y = 5.25$

$x = 5.25 - 2y$

$x + 3y + 3z = 12$

$x + 3y + 3(9.25 - 2x - 2y) = 12$

$-5x - 3y = -15.75$

$5x + 3y = 15.75$

Since $x = 5.25 - 2y$ :

$5(5.25 - 2y) + 3y = 15.75$

$-7y = -10.5$

$y = \frac{10.5}{7}$

$y = 1.5$

Then:

$x = 5.25 - 2y = 5.25 - 2(1.5) = 2.25$

$z = 9.25 - 2x - 2y = 9.25 - 2(2.25) - 2(1.5) = 1.75$

The unit prices are:

$2.25 for a bag of chips.

$1.50 for a liter of pop.

$1.75 for a chocolate bar.

A similar problem, also solved using a system of equations, is given at brainly.com/question/14183076

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2 years ago

Find the highest common factor of 28 and 210 Step by step Please help

Yuri [45]

Hello there!

To find the GCF, we must list all the factors and then just spot the greatest factor.

28: 1, 2, 4, 7, 14, 28.

210 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70.

So 14 is the GCF. Hope this helps you!

~Just a felicitous girlie

#HaveAnAwesomeDay

$SilentNature$

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2 years ago

Rs = 8y + 4 , ST = 4y + 8 , and RT = 36 , find the value of y

Natasha_Volkova [10]

I assume that you meant RS and ST are segments of RT. If that is true then:

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Answer:

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