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lina2011 [118]
3 years ago
11

Please help me out with this so I can keep my kitten.

Mathematics
2 answers:
notka56 [123]3 years ago
5 0

Answer:

y = \frac{32}{3}

Step-by-step explanation:

Firstly, move over the negative 3/4 fraction (don't forget to swap the operation i.e subtract to add):

\frac{y}{8} = \frac{7}{12}  + \frac{3}{4}

Now, to add the two fractions, simply multiply the numerator and denominator by 3:

\frac{3*3}{4*3} = \frac{9}{12}

Now add this to the other fraction:

\frac{9}{12} + \frac{7}{12} = \frac{16}{12}

This can be simplified down by dividing both the numerator and denominator by 4:

\frac{4}{3}

Which now simplifies the original equation to:

\frac{y}{8}  = \frac{4}{3}

Remove the y out of the fraction:

\frac{1}{8}y = \frac{4}{3}

Now multiply both sides by 8:

(\frac{1}{8}y) * 8 = (\frac{4}{3}) * 8

y = \frac{4*8}{3}

y = \frac{32}{3}

Hope this helps!

Terry2 years ago
0 0

Promise it is 32/3.

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Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

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At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0         ......[1]

the solution for the equation is given by:

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

On comparing the given equation with [1] we have;  

a = -16 ,b = 80 and c = 300

then;

t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}

t= \frac{-80\pm \sqrt{6400+19200}}{-32}

t= \frac{-80\pm \sqrt{25600}}{-32}  

Simplify:

t = -\frac{5}{2} = -2.5 sec and t = \frac{15}{2} = 7.5 sec

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8 0
3 years ago
How do you count by 8s and 7s
galben [10]

Answer:

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7x1=7

8x3=24        (8+8+8=24)

7x3=21         (7+7+7=21)

etc.


Hope it helps!

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3 years ago
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2 years ago
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hope this helps!

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