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3 years ago

The sum of two numbers is 84 one number is three times the other find the numbers

2 answers:

6 0

X + y = 84
x = 3y

Plug in 3y for x

(3y) + y = 84
4y = 84
4y/4 = 84/4
y = 21

x = 3y
x = 3(21)
x = 63

x = 63, y = 21

hope this helps

bija089 [108]3 years ago

5 0

It is 63 and 21 because 21x3=63

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3 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

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we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

case A) $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> the term is not a perfect cube

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

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The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

case B) $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> the term is not a perfect cube

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

case C) $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> the term is not a perfect cube

$a^{22}$ ------> the term is not a perfect cube

$b^{16}$ ------> the term is not a perfect cube

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

case D) $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> the term is not a perfect cube

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

case E) $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

Read 2 more answers

Find the square root of the following using division method . iv) 36360

sveticcg [70]

Answer:

190.684

Step-by-step explanation:

To find the square root of 36360, we place a bar over the numbers that we want to find their square root in pairs

, 1 $\dot{3}$ $\overline{63}$ $\overline{60}$

, 1 1

2, 9 2 $\overline{63}$

9 261

38, 0 2 60

0 0

380, 6 26000

, 6 22800

3806, 8 320000

, 8 304480

38068, 4 1552000

Which gives 190.684

3 63 60

We select the divisor to be the largest number which when squares will be equal to the largest number on the left which is 1

We have 1 * 1 = 1

We subtract it from the 3 to get 2

We add the 1's to get 2

We bring down the next pair which is 63 to form the next dividend, 263

We look for a number which will multiply itself to get 263, which is 9, we then have 29 × 9 = 261

We continue till we arrive at the vertical paired numbers which is 190.684.

7 0

3 years ago