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3 years ago

Write the pair of fractions 3/10 and 1/2 as a pair of fractions with a common denominator

1 answer:

7 0

The commen denominator for 10 and 2 would be 10. so 3/10 will stay the same ... and now we change the 1/2 to a fraction that has its denominator as 10. so the new fractions would be ... 3/10, and 5/10.

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Bill and Amy want to ride their bikes from their neighborhood to school which is 14.4 km away. It takes Amy 40 minutes to arrive

vfiekz [6]

Answer:

Amy is (21.6 - 14.4) 7.2 km/hr faster than Bill.

Distance = speed / time

Amy's speed

40 minutes = 40/60 = 2/3 hours

14.4 = speed / 2/3

Speed = 14.4 * 3/2 = 21.6 km/ hr

Bill speed

60 minutes = 1 hour

Speed = 14.4 km / hr.

4 0

3 years ago

Which expression is equivalent to −6(23−12)?

Softa [21]

Answer:

Step-by-step explanation:

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3 years ago

Help me answer this.

Hitman42 [59]

Not sure about this but, is it a line?

7 0

3 years ago

What is the answer to <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bz-5%7D%7Bz%2B1%7D%3D%5Cfrac%7B3%7D%7B2%7D" id="TexFormula1" ti

rusak2 [61]

The correct answer is Z=-13

7 0

2 years ago

Read 2 more answers

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago