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abruzzese [7]
3 years ago
11

How can tell before you divide 387 by 4, that the first digit of the quotient in in the tens place?

Mathematics
1 answer:
iVinArrow [24]3 years ago
7 0
You can tell before you divide 387 by 4, that the first digit of the qoutient is going to be in tens place, because 3 can't go into 4.
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If we divide the numerator and denominator of (6/8) by 2, will its value be changed?
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Step-by-step explanation:

6/8 in simplest form is 3/4 but value is still the same so

1. no

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If YX=YZ, X is 9x-23 and Z is 4x+9 find Z?
Zina [86]

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3 years ago
Someone plz help me with this asap!!!!!
Bas_tet [7]

Answer:

1.) x=8\sqrt{3};y=16

2.) x=1;y=\frac{\sqrt{3}}{2}

3.) x=28 ; y=14\sqrt{3}

4.) x=24 ; y=12\sqrt{3}

5.) x=4\sqrt{3};y=8\sqrt{3}

6.) x=\frac{8\sqrt{3}}{3};y=\frac{16\sqrt{3}}{3}

Step-by-step explanation:

Use the 30°-60°-90° formulas:

a. longer leg=\sqrt{3}*shorter leg

b. hypotenuse=2*shorter leg

1.) Insert values for a:

x=\sqrt{3}*8

Simplify:

x=8\sqrt{3}

Insert values for b:

y=2*8

Simplify:

y=16

2.) Insert values for a:

y=\sqrt{3}*\frac{1}{2}

Simplify:

y=\frac{\sqrt{3}}{2}

Insert values for b:

x=2*\frac{1}{2}

Simplify:

x=1

3.) Insert values for a:

y=\sqrt{3}*14

Simplify:

y=14\sqrt{3}

Insert values for b:

x=2*14

Simplify:

x=28

4.)Insert values for a:

y=\sqrt{3}*12

Simplify:

y=12\sqrt{3}

Insert values for b:

x=2*12

Simplify:

x=24

5.) Insert values for a:

12=\sqrt{3}*x

Divide both sides by \sqrt{3} and rationalize:

\frac{12}{\sqrt{3}}=\frac{\sqrt{3}*x}{\sqrt{3}}\\\\\frac{12}{\sqrt{3}}=x\\\\\frac{\sqrt{3}}{\sqrt{3}}*\frac{12}{\sqrt{3}}\\\\\frac{12\sqrt{3}}{\sqrt{9}}\\\\\frac{12\sqrt{3}}{3}\\\\4\sqrt{3}=x

Flip:

x=4\sqrt{3}

Insert values for b:

y=2*4\sqrt{3}

Simplify:

y=8\sqrt{3}

6.) Insert values for a:

8=\sqrt{3}*x

Divide both sides by \sqrt{3} and rationalize:

\frac{8}{\sqrt{3}}=\frac{\sqrt{3}*x}{\sqrt{3}}\\\\\frac{8}{\sqrt{3}}=x\\\\\frac{\sqrt{3}}{\sqrt{3}}*\frac{8}{\sqrt{3}}\\\\\frac{8\sqrt{3}}{\sqrt{9}}\\\\\frac{8\sqrt{3}}{3}=x

Flip:

x=\frac{8\sqrt{3}}{3}

Insert values for b:

y=2*\frac{8\sqrt{3}}{3}

Simplify:

y=\frac{16\sqrt{3}}{3}

Finito.

5 0
3 years ago
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