Plz help me answer these i dont have a protractor x (Q 7)

Which geometric shape can be described as a fixed distance along a line?

Nina [5.8K]

Perpendicular distance i believe

5 0

3 years ago

Tiana's math test score are 85 and 91 . She has one more test left. In order to earn and A, her average test score must be at le

Zanzabum

85 + 91 + x ÷3 ≥94

You didn't give any choices but this represents the problem.

3 0

3 years ago

A triangle is formed from the points L(-3, 6), N(3, 2) and P(1, -8). Find the equation of the following lines:

Dima020 [189]

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

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

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

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

7 0

3 years ago

Need help fast Thank

ollegr [7]

What you need help with

3 0

3 years ago

Can someoneee please help me

icang [17]

9514 1404 393

Answer:

37°

Step-by-step explanation:

The diagonals of a rhombus are angle bisectors. Angle 1 matches the other half of angle L.

∠1 = 37°

__

Angle 2 is the complement, 53°, and angle 3 is the same as angle 2, 53°.

The diagonals of a rhombus are perpendicular bisectors of each other, so angle 4 is 90°.

7 0

3 years ago