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3 years ago

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Juan's class is going to construct an outdoor garden. The garden will be in the shape of a square, Juan's teacher gave the class

three
options for the area of the garden: 30 square feet, 40 square feet, or 50 square feet.
Part A
Without using a calculator, approximate the side length, to the nearest tenth of a foot, for the garden with an area of 30 square fet. Show
your work.
Part B
The other two garden options have approximate side lengths of 6.3 feet and 7.1 feet. Locate and graph the three points on a horizontal
number line to show the approximation of the side length for each option.

1 answer:

Brrunno [24]3 years ago

7 0

Answer:

Part A: 5.5

Part B: Kindly refer to the attached image for the number line representation.

Step-by-step explanation:

Given that:

Possible area of the first garden = 30 sq ft

Possible area of the second garden = 40 sq ft

Possible length of the second garden = 6.3 ft

Possible area of the third garden = 50 sq ft

Possible length of the third garden = 7.1 ft

To find:

Part A: Side length of the square with area 30 sq ft to the nearest tenth.

Part B: Locating and graphing the three points on a horizontal number line.

Solution:

Formula for area of a square:

$Area =(Side)^2$

Part A: Given that area = 30 sq ft

Putting in the formula to find the value:

$30=Side^2\\\Rightarrow Side = 5.477 \approx \bold{5.5\ ft}$

Part B:

Kindly refer to the attached image for the number line representation of the given two lengths and the length calculated for the first square.

The three lengths are 5.5, 6.3 and 7.1 respectively.

The three numbers to located on the graph are = 5.5, 6.3 and 7.1

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Answer:

Part A) $y=\frac{3}{4}x-\frac{1}{4}$

Part B) $y=\frac{2}{7}x-\frac{5}{7}$

Part C) $y=\frac{2}{7}x+\frac{8}{7}$

see the attached figure to better understand the problem

Step-by-step explanation:

we have

points L(-3, 6), N(3, 2) and P(1, -8)

Part A) Find the equation of the median from N

we Know that

The median passes through point N to midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment NM

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

N(3, 2) and M(-1,-1)

substitute the values

$m=\frac{-1-2}{-1-3}$

$m=\frac{-3}{-4}$

$m=\frac{3}{4}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$

$point\ N(3, 2)$

substitute

$y-2=\frac{3}{4}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{3}{4}x-\frac{9}{4}$

$y=\frac{3}{4}x-\frac{9}{4}+2$

$y=\frac{3}{4}x-\frac{1}{4}$

Part B) Find the equation of the right bisector of LP

we Know that

The right bisector is perpendicular to LP and passes through midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 3

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

so

$m_2=\frac{2}{7}$

step 4

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ M(-1,-1)$ ----> midpoint LP

substitute

$y+1=\frac{2}{7}(x+1)$

step 5

Convert to slope intercept form

Isolate the variable y

$y+1=\frac{2}{7}x+\frac{2}{7}$

$y=\frac{2}{7}x+\frac{2}{7}-1$

$y=\frac{2}{7}x-\frac{5}{7}$

Part C) Find the equation of the altitude from N

we Know that

The altitude is perpendicular to LP and passes through point N

step 1

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 2

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

so

$m_2=\frac{2}{7}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ N(3,2)$

substitute

$y-2=\frac{2}{7}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{2}{7}x-\frac{6}{7}$

$y=\frac{2}{7}x-\frac{6}{7}+2$

$y=\frac{2}{7}x+\frac{8}{7}$

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Hope you all have a great day!

ArbitrLikvidat [17]

u tooooooo!!

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