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tiny-mole [99]
3 years ago
6

Kypton finished 3/8 of his homework. Paxton finished 6/8 of his homework. How much more homework did Paxton do than Kypton?

Mathematics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

Paxton did 3/8 more homework than Kypton.

Step-by-step explanation:

You have to subtract 3/8 from 6/8, after you're done you'll get 3/8. That's the answer.

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What is the least common multiple (LCM) of 5 and 12?
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The least common multiple of 5 and 12 is 60.

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A section of a rectangular floor is covered with square floor tiles as shown below each square tile has a side length of 1/3 foo
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Need more information

Step-by-step explanation:

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4 years ago
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The coordinates of Point S are (2/5, 9 1/8). The coordinates of Point T are (-5 7/10, 9 1/8). What is the distance between Point
m_a_m_a [10]

something noteworthy, the y-coordinate for each point is the same, 9⅛, that means is a horizontal line, over which the x-coordinates are at, so since it's a horizontal line, all we need to do is find, what's the distance between \bf \frac{2}{5}\textit{ and }-5\frac{7}{10}

of course, let's firstly convert the mixed fraction to improper fraction and then check their difference.

\bf \stackrel{mixed}{5\frac{7}{10}}\implies \cfrac{5\cdot 10+7}{10}\implies \stackrel{improper}{\cfrac{57}{10}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{2}{5}-\left[-\cfrac{57}{10} \right]\implies \cfrac{2}{5}+\cfrac{57}{10}\implies \stackrel{\textit{using the LCD of 10}}{\cfrac{(2)2+(1)57}{10}}\implies \cfrac{4+57}{10}\implies \cfrac{61}{10}\implies 6\frac{1}{10}

4 0
3 years ago
Read 2 more answers
A ball is launched upward from a height 40 feet above ground level. The ball’s height at t seconds is given by -16t^2=128=40 .
Nesterboy [21]

Answer:

The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec

Step-by-step explanation:

<u><em>The correct question is</em></u>

A ball is launched upward from a height 40 feet above ground level. the ball's height at t seconds is given by h(t)=-16t^2+128t+40 At what time(s) will the ball be at 100 feet?

we have

h(t)=-16t^{2}+128t+40

so

For h(t)=100 ft

substitute in the equation and solve for x

-16t^{2}+128t+40=100

-16t^{2}+128t-60=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-16t^{2}+128t-60=0

so

a=-16\\b=128\\c=-60

substitute in the formula

x=\frac{-128(+/-)\sqrt{128^{2}-4(-16)(-60)}} {2(-16)}

x=\frac{-128(+/-)\sqrt{12,544}} {-32}

x=\frac{-128(+/-)112} {-32}

x=\frac{-128(+)112} {-32}=0.5

x=\frac{-128(-)112} {-32}=7.5

therefore

The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec

7 0
3 years ago
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