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2 years ago

Shelly ran 2 1/4 miles Lynne ran 3 1/8 Miles WhobRan Farther and by how much

1 answer:

6 0

Lynne ran farther and by 7/8 miles

You can tell Lynne ran further because she ran 3 whole miles while shelly only ran 2 whole miles. Now you have to subtract to see how much Lynne had ran more.

You found a common denominator between 8 and 4. 4 can turn into 8. so you multiply 4x2 and 1x2 to get a mixed number of 2 2/8

Now you have to turn 3 1/8 because you cannot subtract 2/8 from 1/8. You can turn the fraction into 2 9/8 because there are 8 pieces in a whole.

2 9/8 - 2 2/8 = 0 7/8

Hope this helps! :)

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Belle ran 5 5/6 miles around the track. Each lap is 5/6 of a mile. How many laps did Belle run?

STALIN [3.7K]

Belle ran 4 31/36laps

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2 years ago

What is 67 percent as a fraction

olya-2409 [2.1K]

The answer will be 67/100.

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2 years ago

Cubed root x cubed root x2

Yuki888 [10]

Answer:

Final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

Step-by-step explanation:

Given problem is $\sqrt[3]{x}\cdot\sqrt[3]{x^2}$ .

Now we need to simplify this problem.

$\sqrt[3]{x}\cdot\sqrt[3]{x^2}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}$

Apply formula

$\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}$

so we get:

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$

Hence final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

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2 years ago

Help with Pre Calculus

marshall27 [118]

Answer:

The first option

Step-by-step explanation:

The domain of a rational function should be all real numbers except for when the denominator is equal to 0. To find when the denominator is equal to 0 you simply need to find the zeroes of the denominator... but in this case you can do that through factoring and using the quadratic equation.

So first step is going to be to factor out the GCF, which in this case is x. This gives you the equation. $x(2x^2-x-15)$ . So one of the zeroes is when x=0. Now to find the other two zeroes you can use the quadratic equation which is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\$ . So to find the other zeroes you simply plug the values in. a=2, b=-1, c=-15

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-15)}}{2(2)}\\\\x=\frac{1\pm\sqrt{1-4(2)(-15)}}{4}\\\\x=\frac{1\pm\sqrt{121)}}{4}\\\\x=\frac{1\pm11}{4}\\\\x=\frac{12}{4}\\x=3\\\\x=\frac{-10}{4}\\x=-\frac{5}{2}$

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1 year ago

Rewrite the equation 8x + 6x^2 -7=x^2 in standard form and identify a, b, and c.

Vladimir79 [104]

Answer:

$5x^2+8x-7=0\\a=5\\b=8\\c=-7$

Step-by-step explanation:

The standard form of a quadratic is $ax^2+bx+c=0$ . You need to subtract the x^2 from the right to set the quadratic equal to 0. Then a,b, and c are equal to the values of the coefficients in front of the terms.

If this helped, a brainliest would be greatly appreciated!

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2 years ago