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lisabon 2012 [21]
3 years ago
7

There are 6 animals in the pen some are quails and some are goats if there are 20 legs in all how many quails and how many goats

are there?
Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

4 goats

Step-by-step explanation:

Let us represent

Number of quails = x

Number of goats = y

Quails have 2 legs

Goats have 4 legs

Hence:

There are 6 animals in the pen some are quails and some are goats

x + y = 6...... Equation 1

x = 6 - y

if there are 20 legs in all how many quails

2x + 4y = 20..... Equation 2

We substitute 6 - y for x in Equation 2

2(6 - y) + 4y = 20

12 - 2y + 4y = 20

Collect like terms

- 2y + 4y = 20 - 12

2y = 8

y = 8/2

y = 4 goats

The number of goats there are 4 goats

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Multiply each set of numbers and match it with it’s product
Fynjy0 [20]

Answer:

1. 1. (\frac{5}{16}) (-2) (-4) (\frac{-4}{5})= -2 \\2. (2\dfrac{3}{5}) (\frac{7}{9})=(\frac{91}{45})\\3. (\frac{2}{3})(-4)(9)= -24\\4. (\frac{-3}{4}) (\frac{7}{8})=(\frac{-21}{32})

Step-by-step explanation:

These are simple multiplication questions of fractions. We multiply numerator with numerators and denominators with denominators. If numerator and denominator is both divisible by same number we can divide them for simplification.

1. (\frac{5}{16}) (-2) (-4) (\frac{-4}{5})

=(\frac{-10}{16}) (-4) (\frac{-4}{5})\\=(\frac{40}{16}) (\frac{-4}{5})\\=(\frac{-160}{80}) \\dividing\,\, numerator\,\, and\,\, denominator\,\, by\,\, 80\,\,\\= -2

2. (2\dfrac{3}{5}) (\frac{7}{9})

Converting mixed form into fraction form,

=(\frac{13}{5})(\frac{7}{9})

Multiplying numerators with each other and denominators with each other

=(\frac{13*7}{5*9})\\=(\frac{91}{45})

3. (\frac{2}{3})(-4)(9)

Multiplying these terms with each other:

=(\frac{2*-4}{3})(9)\\=(\frac{-8}{3})(9)\\=(\frac{-8*9}{3})\\=(\frac{-72}{3})\\

Dividing numerator and denominator with 3

=-24 

4. (\frac{-3}{4}) (\frac{7}{8})\\=(\frac{-3}{4}) (\frac{7}{8})\\=(\frac{-3*7}{4*8})\\=(\frac{-21}{32})

4 0
3 years ago
Which is one of the solutions to the equation 2x^2 - x - 4 = 0
Alex73 [517]

Answer:

x_{1}=\frac{1+\sqrt{33} }{4}\\x_{2}=\frac{1-\sqrt{33} }{4}

Step-by-step explanation:

Using quadratic formula

x=\frac{-b+-\sqrt{b^{2}-4*a*c} }{2*a}

we will have two solutions.

2x^2 - x - 4 = 0

So, a=2   b=-1  c=-4, we have:

x_{1}=\frac{+1+\sqrt{-1^{2}-4*2*-4} }{2*2}\\\\x_{2}=\frac{+1-\sqrt{-1^{2}-4*2*-4} }{2*2}

Finally, we have two solutions:

x_{1}=\frac{1+\sqrt{33} }{4}\\\\x_{2}=\frac{1-\sqrt{33} }{4}

5 0
4 years ago
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