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lisabon 2012 [21]
3 years ago
12

PLEASE SHOW YOUR WORK WITH YOUR ANSWER. Two bags had 50 kilograms of sugar each. After taking out 3 times as much sugar from bag

one than bag two, bag one had half as much sugar left as bag two. How much sugar is left in each bag?
Mathematics
1 answer:
Svetllana [295]3 years ago
4 0

Answer:

Bag 1: 20

Bag 2: 40

Step-by-step explanation:

Let x be the amount taken out of bag 2

Then the amount left in each bag can be written as:

Bag 1: 50-3x

Bag 2: 50-x

Since we know that half of bag 2 is bag 1, that gives us:

50-3x = 1/2(50-x)

-> 50-3x = 25-x/2

Now lets isolate x and solve:

25 = 5x/2

-> 50 = 5x

-> x = 10

So plug x bag in for the original equations:

Bag 1: 50-3x = 50-3(10) = 20

Bag 2: 50-x = 50-10 = 40

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At this point it seems that you just want to identify the values of b and c.

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Solve 0 = (x – 4)2 – 1 by graphing the related function.
fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

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The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

y=\left(x-\:4\right)^2-1

Finding the x-intercept by substituting the value y = 0

so

y=\left(x-\:4\right)^2-1

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\left(x-4\right)^2-1=0

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\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

\mathrm{Solve\:}\:x-4=\sqrt{1}

\mathrm{Apply\:rule}\:\sqrt{1}=1

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\mathrm{Solve\:}\:x-4=-\sqrt{1}

\mathrm{Apply\:rule}\:\sqrt{1}=1

x-4=-1

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So, when y = 0, then x values are 3, and 5.

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The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

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Who bout to do all this
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Answer:

12, 18, 24, 30

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List the numbers between 11 and 34 that can be evenly divided by 3:

12, 15, 18, 21, 24, 27, 30, 33

Then eliminate the odd numbers, obtaining:

12, 18, 24, 30

These are the four busses that Camille could ride to work.

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How can I solve for the logarithm?
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\bf 12^{2x-8}=15\\\\
-----------------------------\\\\
log_{{  a}}\left( x^{{  b}} \right)\implies {{  b}}\cdot  log_{{  a}}(x)\impliedby thus\\\\
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