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3 years ago

2/3 and 5/6 common denominator

1 answer:

7 0

The common denominator would be 6. Since, the least common multiple of 3 and 6 is 6.
3:
3, 6, 9
6:
6, 12, 18
Least common multiple is 6

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3 is 12% of please help asap

Korvikt [17]

Answer:

25%

Step-by-step explanation:

hopes this helps you

8 0

3 years ago

Read 2 more answers

HELP PLEASE!!!!!! Anaya bought a sweater that was on sale. The sweater originally cost $34.50 but was purchased for $27.60. What

Elan Coil [88]

Answer: 80%. Hope this helps, please consider making me Brainliest.

Step-by-step explanation:

To find the percentage, divide the sale cost by the original cost:

27.60/34.50

Let's multiply both sides by 10 to make the operation easier:

27.6/34.5 ( I eliminated the zero's because they kind of have no use) -->

276/345, now solve:

276/345 = 0.8

0.8 = 80%

The percentage is 80%.

5 0

1 year ago

HELP PLEASE ASAP IM ON THE LAST ONE

inna [77]

Answer:

y=9; z=16

Step-by-step explanation:

7 0

3 years ago

given the graph of the line represented by the equation f(x)=-2x+b,if b is increased 4 units, the graph of the new line would be

jek_recluse [69]

Increasing function value shifts the line up.
answer: 2. up

6 0

2 years ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago