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3 years ago

6

A camping ground began with 240 pounds of food and each day consumed about 12 pounds. After how many days did they have 180 poun

ds left

2 answers:

Mice21 [21]3 years ago

7 0

If you take 240 divided by 12 = 20
Then take 180 divided by 12 = 15
Take 20-15=5 after 5 days they had 180lbs left

mash [69]3 years ago

3 0

Answer:

5 days

Step-by-step explanation:

Set up linear equation; food= 240-12(day) f=240-12d

f(1)=240-12(1)=240-12=228

f(2)=240-12(2)=240-24=216

Now set food to equal 180, solve for days:

180=240-12d

-240 -240

-60=-12d

/-12 /-12

5=d

To check: f(5)=240-12(5)=240-60=180

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Answer:

1. The ratio of their diameters is 8/5 = 8 : 5

2. The ratio of their surface area is (8/5)²

3. The ratio of their volume is (8/5)³

4.The area of the base of the larger cylinder is 128 cm²

Step-by-step explanation:

Given that the two cylinders are similar, we have;

Two cylinders are similar when the ratio of their heights is equal to the ratio of their radii

Therefore, we have;

1. The ratio of the height of the two cylinders = 8/5 = The radio of their radii = r₁/r₂

The ratio of their diameter = D₁/D₂ = 2·r₁/2·r₂ = r₁/r₂ = 8/5

The ratio of their diameters D₁/D₂ = 8/5 = 8 : 5

2. The surface area of the cylinders = 2·π·r·h + 2·π·r²

Therefore, we have;

(2·π·r₁·h₁ + 2·π·r₁²)/(2·π·r₂·h₂ + 2·π·r₂²) = (r₁·h₁ + r₁²)/(r₂·h₂ + r₂²)

h₁ = h₂ × 8/5

r₁ = r₂ × 8/5

= (8/5)²(r₂·h₂ + r₂²)/(r₂·h₂ + r₂²) = (8/5)²

The ratio of their surface area = (8/5)²

3. The volume of the cylinder = π·r²·h

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The ratio of their volume = (8/5)³

4. The ratio of the area of the base of the larger cylinder to the area of the base of the smaller cylinder is (8/5)²

Therefore if the area of the base of the smaller cylinder is 50 cm², the area of the base of the larger cylinder = 50 cm² × (8/5)² = 128 cm²

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Learning Task 3. Find the equation of the line. Do it in your notebook.

Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( $m = 5$ , $x = -1$ , $y = 4$ )

$4 = (5)\cdot (-1)+b$

$4 = -5 +b$

$b = 9$

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

$m = \frac{2}{5}$

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( $m = 3$ , $b = 4$ )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

$b = 6$ (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

$m = 2$

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

$-m+b = -2$ (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

$b = -2+m$

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

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2 years ago