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umka21 [38]
3 years ago
12

Caleb and Suzanne each have a pile of blue and white chips. Caleb determines in his pile,three out of every four chips is blue.

Suzanne determines in her pile, one out of every five chips are blue. How many total chips does each student have if Caleb and Suzanne each have 6 blue chips?
Mathematics
1 answer:
Artist 52 [7]3 years ago
8 0

Answer:

Total number of chips Caleb have = 8

Total number of chips Suzanne have = 30

Step-by-step explanation:

In Caleb case -

Three out of every four chips is blue

⇒ If Caleb has total chips = 4

Then Blue chips = 3

⇒White chips = 4 - 3 = 1

Now,

If Caleb have 6 blue chips

⇒3 Blue chips = 4 total chips

  1 Blue chip = \frac{4}{3} total chips

⇒6 blue chips = \frac{4}{3} ×6 = 4×2 = 8 total chips

∴ we get

Number of chips become = 8

In Suzanne case -

One out of every five chips are blue.

⇒ If Suzanne has total chips = 5

Then Blue chips = 1

⇒White chips = 5 - 1 = 4

Now,

If Suzanne have 6 blue chips

⇒1 Blue chips = 5 total chips

⇒6 blue chips = 5×6 = 30 total chips

∴ we get

Number of chips become = 30

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Answer:

Step-by-step explanation:

False

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One sixth of a jar of sweets contains 11 sweets. How many sweets does the jar contain when its full?
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66 sweets

Step-by-step explanation:

all you need to do is multiply 11 by 6 to get your answer seeing that ⅙ of the jar is 11

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A recipe for banana pudding calls for 2/3 cups of sugar for the flour and 1/4 cups of sugar for the meringue topping, how many c
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Answer: 11/12 cups sugar

1 1/4 lbs  candy

Step-by-step explanation:

2/3 + 1/4=8/3 + 3/4=11/12

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6 0
3 years ago
2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?
7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

The solutions to the quadratic equation 2x^2\:-\:5x\:+\:67\:=\:0 are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Step-by-step explanation:

Considering the equation

2x^2\:-\:5x\:+\:67\:=\:0

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

so

\frac{2x^2-5x}{2}=\frac{-67}{2}

x^2-\frac{5x}{2}=-\frac{67}{2}

Lets now solve the equation by completeing the remaining steps

Write equation in the form: x^2+2ax+a^2=\left(x+a\right)^2

Solving for a,

2ax=-\frac{5}{2}x

a=-\frac{5}{4}

\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}

Completing the square

\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

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

Solving

x-\frac{5}{4}=\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}

x-\frac{5}{4}=i\sqrt{\frac{511}{16}}       ∵ Applying imaginary number rule \sqrt{-1}=i

x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}

-\frac{5}{4}=i\frac{\sqrt{511}}{4}

x=\frac{5}{4}+i\frac{\sqrt{511}}{4}

Similarly, solving

x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}    ∵ Applying imaginary number rule  \sqrt{-1}=i

x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Therefore, the solutions to the quadratic equation are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

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