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liubo4ka [24]
3 years ago
8

Anyone know this mathswatch fraction question ???

Mathematics
2 answers:
Lynna [10]3 years ago
8 0

Answer:

a)= 5 \frac{9}{14}  \\ \\ b) = 7 \frac{1}{2}  \\

Step-by-step explanation:

a)4 \frac{1}{7}  + 1 \frac{1}{2}  \\  \frac{29}{7}  +  \frac{3}{2}  \\  \frac{58 + 21}{14}  \\  =  \frac{79}{14}  \\  = 5 \frac{9}{14}

b)4 \frac{1}{2}  \div  \frac{3}{5}  \\  \frac{9}{2}  \div  \frac{3}{5}  \\  \frac{9}{2}   \times  \frac{5}{3}  \\  =  \frac{45}{6}  \\  =  7 \frac{3}{6}  \\  = 7 \frac{1}{2}

Sophie [7]3 years ago
3 0

Answer:

5 9/14

7 1/2

Step-by-step explanation:

a) 4 1/7 + 1 1/2= 4+ 1/7 +1 +1/2 = 5 + 1/7 +1/2 = 5+ 2/14 + 7/14= 5 + 9/14= 5 9/14

b) 4 1/2 ÷ 3/5 = 9/2 ÷ 3/5 = 9/2 *5/3= 3/2*5= 15/2= 7 1/2

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Given the equation p = s Subscript 1 Baseline t minus 2 Subscript 2 Baseline t, which equation is solved for t? t = p (s Subscri
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Answer:

Option C) is correct

That is t=StartFraction p Over s Subscript 1 Baseline minus s Subscript 2 Baseline EndFraction

It also can be written as t=\frac{p}{s_{1}-s_{2}}

Step-by-step explanation:

Given equation can be written as below:

p=s_{1}t-s_{2}t

Now to solve the equation for t:

p=s_{1}t-s_{2}t

Taking common term t outside on RHS we get

p=(s_{1}-s_{2})t

\frac{p}{s_{1}-s_{2}}=t

Rewritting the above equation as below

t=\frac{p}{s_{1}-s_{2}}

Therefore option C) is correct

That is t=StartFraction p Over s Subscript 1 Baseline minus s Subscript 2 Baseline EndFraction

It also can be written as t=\frac{p}{s_{1}-s_{2}}

5 0
3 years ago
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Step-by-step explanation:

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2 years ago
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Write a two column proof 2x+6 and 96 prove: x=45
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You spent 1/2 of your money at the theater and 1/4 at an arcade. you have $17.50 left. How much money did you originally have?
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Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4

first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
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subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x=\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a} so

1x^2-2x+4=0
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b=-2
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x=\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}
x=\frac{ 2+/-\sqrt{4-16} }{2}
x=\frac{ 2+/-\sqrt{-12} }{2}
we have \sqrt{-12} and that doesn't give a real solution
therefor there are no real solutions
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i=\sqrt{-1}
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x=1-i\sqrt{3} or x=1+i\sqrt{3} (those are the 2 numbers)

 





6 0
3 years ago
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