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3 years ago

If a baker doubles a recipe that calls for 6-2/3 cups of flour. How many cups will be needed in all?

Mathematics

2 answers:

Musya8 [376]3 years ago

6 0

The baker would need 13 &1/3 cups.

nadya68 [22]3 years ago

4 0

Hey there!

The easiest way I could think to do this is by converting your mixed number to an improper fraction and multiplying the fraction by 2, or 2 over 1.

$6 \frac{2}{3} = 6*3+2 = \frac{20}{3}$

$\frac{20}{3} * \frac{2}{1}$

To multiply fractions, you can just multiply the numerators and denominators and simplify, if applicable.

$\frac{20}{3} * \frac{2}{1} = \frac{40}{3}$

Since you can't simplify this fraction in its improper form, just convert it back into a mixed number.

$\frac{40}{3} = 13 \frac{1}{3}$

So, your answer will be $13 \frac{1}{3}$ .

Hope this helped you out! :-)

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Divide x^3-3x^2-10x+24 by x-1

lesya [120]

Answer:

x^2 - 2x - 12 with remainder 12

Step-by-step explanation:

Synthetic division is the fastest way in which to carry out this division.

The divisor (x - 1) from long division corresponds to the divisor 1 in synthetic division. Setting up synthetic division, we get:

1 / 1 -3 -10 24

1 -2 -12

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1 -2 -12 12

The first three digits {1, -2, -12} are the coefficients of the quotient, and 12 represents the remainder:

The quotient is 1x^2 - 2x - 12 and the remainder is 12.

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3 years ago

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Step-by-step explanation:

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2 years ago

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2 – 3х + 5х = 16 — 8х + 8x

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Answer:

2 – 3х + 5х = 16 — 8х + 8x

2 - 8x = 16

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2 years ago

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What is the length of the missing base of the trapezoid?

Eddi Din [679]

Answer:

6 inches

Step-by-step explanation:

Area = (base1 + base2)/2 · height

base1 = b

base2= 8

height = 3

Area = 21

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3 years ago

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

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