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3 years ago

11

Joseph has 150 animals on his farm. ⅕ of the animals are chickens, ⅕ of the animals are cows, and the rest of the animals are go

ats. How many goats does Joseph have on his farm?

2 answers:

ASHA 777 [7]3 years ago

4 0

Answer:

90 goats

Step-by-step explanation:

Number of chickens in farm = (1/5) of 150

= $\frac{1}{5}*150$

= 1 * 30

= 30

Number of cows in farm = $\frac{1}{5}*150$

= 1 * 30

= 30

Number of goats in farm = 150 - (30 + 30)

= 150 - 60

= 90

Gennadij [26K]3 years ago

3 0

Answer:

Joseph have 90 goats on his farm

Step-by-step explanation:

To solve this problem we first have to subtract the fraction of chickens and cows from 1 to find the fraction that represents goats

(1 - ⅕) - ⅕ = 3/5

Now that we know the fraction of goats we multiply it by the total number of animals

3/5 * 150 = 90

Joseph have 90 goats on his farm

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What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

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Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

$y=\dfrac{4}{2}$

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

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