All categories

3 years ago

Find the equivalent fraction of 2/3 having denominator 6 and 6 and numerator 10.

Mathematics

1 answer:

4vir4ik [10]3 years ago

7 0

Step-by-step explanation:

having denominator 6,

X/6= 2/3

X= 4

so, 4/6= 2/3

having numerator 10,

10/X= 2/3

X= 15

so, 10/15= 2/3

You might be interested in

Couch Emanuel is trying to lose weight. he starts 2017 weighing 296 pounds. If he plans to burn enough calories to lose 3 pounds

Fantom [35]

Answer:

15 weeks

Step-by-step explanation:

I just answered this one not too long ago

8 0

3 years ago

The figure is made up of 3 semicircles. Find its perimeter in terms of pi.

icang [17]

Answer:

The answer is 20pi

Step-by-step explanation:

perimeter of a semicircle = pi* radius

perimeter of first semicircle = 5.pi

perimeter of second semicircle = 5.pi

perimeter of third semicircle = 10. pi

total perimeter = 5pi +5pi + 10pi

= 20pi

5 0

3 years ago

4x - 1 = -13 show the work please

m_a_m_a [10]

Answer:

x = -3

Step-by-step explanation:

4x - 1 = -13

add 1 to each side

4x = -12

x = -3

5 0

3 years ago

Read 2 more answers

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers

Which of the following congruency statements is correct? Yzx = bcd, yzx = dcb, zxy =dcb, yxz = cdb

goldenfox [79]

Answer:c

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the equivalent fraction of 2/3 having denominator 6 and 6 and numerator 10.​

Find the equivalent fraction of 2/3 having denominator 6 and 6 and numerator 10.