3 years ago

How are common denominators and common multiples the same

1 answer:

7 0

Find the Least Common Multiple of the denominators (which is called the Least Common Denominator). Change each fraction (using equivalent fractions) to make their denominators the same as the least common denominator. Then add (or subtract) the fractions, as we wish!

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A CD usually sells for $14 if the CD is 30% off and still taxes 8% what is the total price of the CD including tax

lesya692 [45]

The CD will cost $10.584
The 0.004 will likely round up to $10.59

4 0

4 years ago

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(3rt^2)^2 (9rs^2t^3)^-2

MakcuM [25]

$( {3}^{2} {r}^{2} {t}^{4} )( \frac{1}{ {9}^{2} {r}^{2} {s}^{4} {t}^{6} } )$
$\frac{1}{9 {s}^{4} {t}^{2} }$

4 0

3 years ago

When multiplying 1,729 times 308, does Clayton got a product of 53,253. Without calculating, does his product seem reasonable? E

ExtremeBDS [4]

Nope if you round the first number to 2000 and the second to 300 you should expect around 600000 which is not the case.

3 0

4 years ago

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Select whether the equation has a solution or not.

vredina [299]

Answer: The equation has a solution .

Step-by-step explanation:

Since we have given that

$\sqrt{x+1}=7-2\sqrt{x}$

Squaring on both the sides, we get that,

$x+1=(7-2\sqrt{x})^2\\\\x+1=49+4x-28\sqrt{x}\\\\3x+48-28\sqrt{x}=0$

Let $\sqrt{x}=y\implies x=y^2$

So, our equation becomes,

$3y^2-28y+48=0\\\\y=\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3}\\\\\sqrt{x}=\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3}\\\\x=(\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3})^2$

So, $x=\dfrac{248}{9}\pm \dfrac{56\sqrt{13}}{9}$

Hence, the equation has a solution .

4 0

3 years ago

How would I solve -(28-45)=17

Nonamiya [84]

-(28-45)
= -28 -(-45)
=-28 + 45
= 45 - 28
= 17

hope it helps

5 0

4 years ago