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3 years ago

a farmer notes that in a field of goats and geese there are 41 heads and 114 feet how many of each animal are there

Mathematics

1 answer:

solniwko [45]3 years ago

6 0

Answer:

There are 25 geese and 16 goats

Step-by-step explanation:

Let O represent the # of goats and E the # of geese. Remembering that goats have four feet and geese only two, and that each animal has only one head, we can write the following equations to determine O and E:

Number of heads: 41; this is the total of the # of goat heads and the # of goose heads: O + E = 41

(4 feet per goat)O + (2 feet per goose)E = 114 (There are a total of 114 feet)

Then we solve the following system of linear equations:

O + E = 41

4O + 2E = 114

Multiplying the first equation by -2 results in -2O - 2E = -82.

Combining this result with the second equation, we obtain the system

4O + 2E = 114

-2O - 2E = -82

-------------------------------

2O = 32

Then O = # of goats = 32/2, or 16, and so the equation O + E = 41 becomes

E = # of geese = 41 - O (remembering that we found O to be 16), or

E = 41 - 16 = 25

There are 25 geese and 16 goats.

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inysia [295]

Answer:

12.699

Step-by-step explanation:

i think if you do 26-22.4, you get the leg of the small triangle which is 3.6

then do Pythagorean theorem

3.6^2 + x^2 = 13.2^2

12.96 +x^2 = 174.24

174.24-12.96

x^2 = 161.28

take square root

x = 12.699

6 0

2 years ago

Which numbers are solutions of x2+20=9x+2? Choose exactly 2 awnsers. 2 3 5 6

Dahasolnce [82]

Answer:

my answer is gonna be 2 and 5

7 0

3 years ago

Read 2 more answers

Write a rule for the nth term of the arithmetic sequence: a11= 50, d = 7

ivolga24 [154]

Answer:

Required rule for $n^{th}$ is $a_{n}=7n-27$ .

Step-by-step explanation:

Given that,

$a_{11} = 50,\ \ d=7$

From the question: we have to write the $n^{th}$ term of Arithmetic sequence.

Arithmetic Sequence or Arithmetic progression (A.P) : It is a sequence which possess that difference between of two successive sequence is always constant.

$a_{1} ,a_{2},a_{3},a_{4}.....................a_{n-1},a_{n}$

where, $a_{1}$ is the first term of A.P

$d$ is the common difference.

$a_{n}$ is the last term or general term.

The above sequence to be in A.P then their common difference should be equal.

$d=a_{2}-a_{1} =a_{3}-a_{2}=a_{4} -a_{3} ..........................a_{n}-a_{n-1}$

Now, Formula of General Term is $a_{n}=a+(n-1)d$

So, $a_{11}= a+(11-1)d\\ a_{11} = a+10d$

Substituting the value of $a_{11} = 50,\ \ d=7$ we get,

$50=a+10\times7\\50=a+70\\a=-20$

Then General term ( $a_{n}$ ) of given data is

$a_{n}=-20+(n-1)7\\a_{n}=-20+7n-7\\a_{n}=7n-27$

Therefore, Required rule for $n^{th}$ is $a_{n}=7n-27$ .

5 0

3 years ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$