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JulijaS [17]
4 years ago
5

Helpers are needed to prepare for the fete. Each helper can make either 2 large cakes or 35 small cakes per hour. The kitchen is

available for 3 hours and 20 large cakes and 700 small cakes are needed. How many helpers are required?
Mathematics
1 answer:
NeX [460]4 years ago
5 0

Answer:

You require 10 helpers

Step-by-step explanation:

The problem could be solved in many ways ( like using an optimization software) but I propose you this.

Start with production of large cakes.  In the time kitchen is available one helper could make, if only works in large cakes:

(2 large cakes / 1 hour) * (3 hours ) = 6 large cakes

So with 3 helpers using all their 3 hours we would have

3 helpers *6 large cakes /helper  = 18 large cakes.

We need 2 more large cakes so we can use one more helper and in his first hour he can produce

(2 large cakes / 1 hour) * (1 hour ) = 2 large cakes

The 4th helper has two hours of work left so he can produce small cakes.

(35 small cakes/hour* 2 hours = 70 small cakes.

With 4 helpers we have the 20 large cakes and 70 small ones. We still have to makes

700 - 70 = 630 small cakes left

In the 3 hour period a helper can make

(35 small cakes / 1 hour) * (3 hours ) = 105 large cakes

So if 1 helper does  105 large cakes  we would need to finish production with

630 / 105 =  6 helpers

So in total we have 3 helpers in only large cakes 6 in only small cakes and one helper doing both - total 10 helpers

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Juliette [100K]
x-the\ number\\\\x^2-(5+3x)=0\\\\x^2-5-3x=0\\\\x^2-3x-5=0

Use\ the\ quadratic\ formula:\\a^2x+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ then\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solution

x^2-3x-5=0\\\\a=1;\ b=-3;\ c=-5\\\\\Delta=(-3)^2-4\cdot1\cdot(-5)+3+20=29 \ \textgreater \  0\\\\\sqrt\Delta=\sqrt{29}\\\\x_1=\dfrac{3-\sqrt{29}}{2\cdot1}=\dfrac{3-\sqrt{29}}{2} \ \textless \  0\\\\x_2=\dfrac{3+\sqrt{29}}{2\cdot1}=\dfrac{3+\sqrt{29}}{2} \ \textgreater \  0

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6 0
3 years ago
Solve 5x^2 − 3x + 17 = 9.
Vedmedyk [2.9K]

Answer:

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

or

⇒ 5x² − 3x + 17 - 9 = 0

or

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

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

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = \frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}

or

x = \frac{3\pm\sqrt{9-160}}{10}

or

x = \frac{3+\sqrt{-151}}{10} and x = \frac{3-\sqrt{-151}}{10}

or

x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

here i = √(-1)

Hence,

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

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Answer:

Step-by-step explanation:

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WITH replacement of first card:

P(two 10s are drawn) = P(first card is a 10)*P(first card is a 10) = (4/13)(4/13) =

16/169

7 0
3 years ago
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