Unit 4 linear equations homework 9 parallel and perpendicular lines (day1) activity
Directions if segments AB and CD are parallel,perpendicular, or neither
Answers
The lines that are parallel continuously have equal distances from each
other and perpendicular lines form 90° angle at their intersection.
The correct responses are;
Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel. ║Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neitherSegment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neitherThe equations are parallel, ║The equations are perpendicular, ⊥The equations are perpendicular, ⊥The equations are neither parallel or perpendicularReasons:
Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular
[tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel where; Slope of [tex]\overline{AB}[/tex] = Slope of [tex]\overline{CD}[/tex]
[tex]\displaystyle \overline{AB} \ and \ \overline{CD} \ are \ perpendicular\ where; \ Slope \ of \ \overline{AB} = -\frac{1}{Slope \ of \ \overline{CD}}[/tex]
[tex]1. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{\frac{13 - 3}{-2 - 0}} = \frac{10}{-2} = -5[/tex]
[tex]\geq \displaystyle Slope \ of \ \overline{CD} = \frac{3 - 0}{10 - (-5)} = \frac{1}{5}[/tex]
[tex]\displaystyle \mathbf{Slope \ of \ \overline{AB}} = -\frac{1}{Slope \ of \ \overline{CD}}[/tex], [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular
[tex]2. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{\frac{7- 1}{3 - (-6)}} = \frac{6}{9} =\frac{2}{3}[/tex]
[tex]\geq \displaystyle Slope \ of \ \overline{CD} = \frac{-5 - (-1)}{-6 - 0} = \frac{-4}{-6}=\frac{2}{3}[/tex]
Slope of [tex]\mathbf{\overline{AB}}[/tex] = Slope of [tex]\overline{CD}[/tex] , [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel
[tex]3. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{ \frac{2- 4}{-6 - (-2)} = \frac{-2}{-4}} =\frac{1}{2}[/tex]
[tex]\displaystyle Slope \ of \ \overline{CD} = \frac{11 - (-7)}{-1 - 5} = \frac{18}{-6}=-3[/tex]
[tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither parallel or perpendicular
[tex]4. \ \displaystyle Slope \ of \ \overline{AB} = \frac{8- 3}{-3 - 2} = \frac{5}{-5} =-1[/tex]
[tex]\displaystyle Slope \ of \ \overline{CD} = \frac{6 - 2}{-4 - (-8)} = \frac{4}{4}=1[/tex]
[tex]\displaystyle Slope \ of \ \overline{AB} = \mathbf{-\frac{1}{Slope \ of \ \overline{CD}}}[/tex], [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular
[tex]5. \ \displaystyle Slope \ of \ \overline{AB} = \frac{-1 - 2}{-8 - (-4)} = \frac{-3}{-4} =\frac{3}{4}[/tex]
[tex]\displaystyle Slope \ of \ \overline{CD} = \mathbf{\frac{-3 - 6}{0 - 12}} = \frac{-9}{-12}=\frac{3}{4}[/tex]
Slope of [tex]\overline{AB}[/tex] = Slope of [tex]\overline{CD}[/tex] , [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel
[tex]6. \ \displaystyle Slope \ of \ \overline{AB} = \frac{5 - (-1)}{6 - 3} = \frac{6}{3} =2[/tex]
[tex]\displaystyle Slope \ of \ \overline{CD} = \mathbf{\frac{-5 - 7}{2 - (-4)}} = \frac{-12}{6}=-2[/tex]
[tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither parallel or perpendicular
Directions: To determine if the equations are parallel or perpendicular
Solution:
The equations are parallel if when expressed in the form, y = m·x + c, the value of m are equal
The equations are perpendicular, when we have;
[tex]\displaystyle The \ value \ of \ \mathbf{m} \ in \ one \ equation = -\frac{1}{The \ value \ of \ \mathbf{m} \ in \ the \ other\ equation }[/tex]
7. First equation
3·x + 2·y = 6
Therefore; [tex]\displaystyle y = -\frac{3}{2} \cdot x + 3[/tex]
Second equation; [tex]\displaystyle y = \mathbf{-\frac{3}{2} \cdot x + 5}[/tex]
The value of m which is the coefficient of x are both equal to [tex]\displaystyle -\frac{3}{2}[/tex]
Therefore, the equations are parallel
8. Equation 1; 3·y = 4·x + 15
Therefore;
[tex]\displaystyle y = \frac{4}{3} \cdot x + 5[/tex]
Equation 2; 9·x + 12·y = 12
[tex]\displaystyle y = -\frac{9}{12} \cdot x + 1 = 1 - \frac{3}{4} \cdot x[/tex]
The slope of the equation 2 is the negative inverse of the slope of equation 1, therefore, the equations are perpendicular
9. Equation 1: 8·x - 2·y = 4
Therefore;
y = 4·x - 2
Equation 2: x + 4·y = -12
Therefore;
[tex]\displaystyle y = -\frac{1}{4} \cdot x - 3[/tex]
The slope of the equation 2 is the negative inverse of the slope of equation 1, therefore, the equations are perpendicular.
10. Equation 1: 3·x + 2·y = 10
Therefore;
[tex]\displaystyle y = -\frac{3}{2} \cdot x - 5[/tex]
Equation 2: 2·x + 3·y = -3
[tex]\displaystyle y = -\frac{2}{3} \cdot x -1[/tex]
The slopes of the equation 1 and 2 are not equal, and are not the negative inverse of each other, therefore, the equations are neither parallel nor perpendicular.
Learn more about parallel and perpendicular lines here:
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Related Questions
Please help! It's a quiz and I can't check the answer so it needs to be correct!!!
Answers
Answer:
the answer is jesus asdlkfjsfs
20 points im confused with this one
Answers
Answer: D
Step-by-step explanation:
Answer:
i say go with D
Step-by-step explanation:
The Radius of a circle is 5 cm. Find the area of the circle, correct to one decimal place.
Answers
Answer:
78.5
Step-by-step explanation:
[tex]r^{2}[/tex]x3.14 (pie)
so then substitute.
5x5=25
25x3.14=78.5
fill in the blanks ANGLES (I WILL GIVE BRAINLIEST! PLS HELP!)
refer to attachment for question
Answers
Answer:
See belowStep-by-step explanation:
If two triangles are congruent, then the corresponding sides and angles are congruent with each other.
Given:
ΔRED ≅ ΔSUNThe following are congruent parts:
∠E ≅ ∠URE ≅ SU∠DRE ≅ ∠NSU∠R ≅ ∠SNote Pay attention to order of vertices when comparing the congruent parts
[tex]\\ \sf\longmapsto <E\cong <U[/tex]
[tex]\\ \sf\longmapsto \overline{RE}=\overline{SU}[/tex]
[tex]\\ \sf\longmapsto ∆DRE=∆SUN[/tex]
[tex]\\ \sf\longmapsto <R\cong <S[/tex]
Hope it helps
The solution to the equation log2x log2(x â€"" 6) = 4 is x =.
Answers
The solution to the logarithmic equation [tex]\mathbf{log_2(x-6) = 4}[/tex] is x =24
The logarithmic equation is given as:
[tex]\mathbf{log_2(x-6) = 4}[/tex]
Apply exponential law of logarithm
[tex]\mathbf{(x-6) = 2^4}[/tex]
Remove the brackets
[tex]\mathbf{x-6 = 2^4}[/tex]
Add 6 to both sides
[tex]\mathbf{x-6 +6= 6+ 2^4}[/tex]
Express 2^4 as 16
[tex]\mathbf{x-6 +6= 6+ 16}[/tex]
Add -6 and 6
[tex]\mathbf{x= 6+ 16}[/tex]
Add 6 and 16
[tex]\mathbf{x= 24}[/tex]
Hence, the solution to the logarithmic equation [tex]\mathbf{log_2(x-6) = 4}[/tex] is x =24
Read more about logarithmic equation at:
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Angie and her friends are heading to the Glacier Heights snow park. They plan to purchase the Glacier group package, which costs $76 for 8 people. That's $2 less per person than the normal cost for an individual. What is the normal cost for an individual?
Answers
Answer
The normal cost would be $11.50
Number Explanation
1) 76 ÷ 8 = 9.5
2) 9.5 + 2 = 11.5 or 11.50
Word Explanation
First, I divided the total cost 76 by the number of people 8, to find out how much it costs per person. I got 9.5! Next, I added 2 to 9.5 because in the problem it said this is $2 less than the normal cost per person. I got 11.5! That was my answer.
How do I know which region to shade for this inequality
y ≤ 2x + -4?
Answers
Answer:
You look at the sign between y and 2x
Step-by-step explanation:
To find which way to shade, look at your symbol. The symbol you have is a less than or equal to, you shade to the left. Because it has the line underneath, you have a closed circle. (This is for line plots so I don't know if this completely helps)
Hopefully it does help though!!
Which graph shows the lowest cost per square foot for a new home?
Answers
Answer:
the bottom right
Step-by-step explanation:
What is the length of side x?
Answers
Answer:
21
Step-by-step explanation:
10.5 ÷ 3 = 3.5
3.5 × 6 = 21
Graph the line that has a slope of 4 and includes the point (0,4)
Answers
Answer:
Step-by-step explanation:
it goes straight across 4 ( up and down)
Answer:
Step-by-step explanation:
Start with the standard form of an equation for a straight line:
y=mx+b,
where m is the slope and b is the y-intercept (the value of y when x = 0).
We are given the slope (4), so we have:
y = 4x + b
We can calculate b from the given point, (0,4). Just enter the values for x and y and solve for b. But in this case the given point already tells us the value of b since at x = 0, y = 4. So we can enter it directly.
------
For reference, if we went ahead and solved for b, this is what we'd get:
y = 4x + b
4 = 4*(0) + b
b = 4
-----
So the final equation of this line is y = 4x + 4
The graph is attached.
5. Write the following numbers in order from least to greatest. 1.23; 1.2; 2.31; 3.2
Answers
Please help me, I struggle the most in this class and this is the last question
Answers
I have to do the evens- I was out due to epilepsy and have no clue what I'm doing
Worth 50 points
Answers
Step-by-step explanation:
I'm assuming your solving for [tex]x[/tex]
Picture 2.
For this one since the two sides are equal
[tex]2x-12=12\\2x=24\\x=12[/tex]
Picture 4.
We can use the Pythagorean theorem for this one, which is [tex]a^2+b^2=\sqrt{c}[/tex]
[tex]4\cdot4=16\\3\cdot3=9\\16+9=25\\\sqrt{25}=5[/tex]
[tex]x=5[/tex]
Picture 6.
We can use the Pythagorean theorem for this one, which is [tex]a^2+b^2=\sqrt{c}[/tex]
[tex]6\cdot6=36\\9\cdot9=81\\36+81=117\\[/tex]
[tex]\sqrt{117}=approximately[/tex] [tex]10.81[/tex]
Picture 8.
[tex]EG=6[/tex]
Since half of [tex]EG[/tex] is [tex]3[/tex]
[tex]3\cdot2=6[/tex]
[tex]EG=6[/tex]
Brainliest, please :)
x+6=0
3x-2y = 10
simultaneous equations
Answers
3(-6)-2y=10
-18-2y=10
2y=10+18
2y=28
Y=28/2
Y=14
X+6=0
X=0-6
X=-6
3x-2y=10
2y=3x-10
If the side lengths are
doubled, then which of
the following
statements about its
perimeter will be true?
The new perimeter will be
times the old
Answers
Answer:
It would be 2 times the old perimeter.
Solve for x using common denominators.[tex]\frac{3}{4(x+2)}=\frac{x}{(x+2)}[/tex]
[A]-\frac{4}{3}[/tex]
[B]\frac{4}{3}[/tex]
[C]-\frac{3}{4}[/tex]
[D]\frac{3}{4}[/tex]
Answers
Step-by-step explanation:
3/(4(x+2)) = x/(x+2)
the common denominator would have to be 4(x+2).
to get the right side to this denominator, we have to multiply numerator and denominator (top and bottom) by 4.
and then we get
3/(4(x+2)) = (4x)/(4(x+2))
and now as both sides are "standing" on the same denominator, we can simply compare only the top parts.
3 = 4x
and therefore we can solve
x = 3/4
and D is then the right answer choice.
how can u calculate 23x4 if the 4 buton is broken?
Answers
Answer:
23x2x2
Step-by-step explanation:
You jog 2 kilometers in 12 minutes. Your friend jogs 3 kilometers in 16.5 minutes. Who jogs faster?
Answers
Answer:
your friend is faster.
Step-by-step explanation:
2 km in 12 min means 3 km in 18 min.
you friend jogs those same 3 km in 16.5 min, so he or she is faster.
what does n mean in -10(n-5)=40
Answers
-10n - 50 = 40
Add 50 to both sides
-50 + 50 would cancel
- 50 + 40 = -10
Divide by -10
-10/-10 = 1
N=1
Answer:
Divide both sides by -10
-10n+2=40
subtract 2 from both sides so
40-2=38
10n/10= 38/10
3.8 = n
Step-by-step explanation:
Can someone please solve this problem? I’ll mark you at brainlest! (I mostly need help with 10 and down but you can do whatever!)
Answers
Answer:
10. 45/6
Step-by-step explanation:
Whatever you do to the top you have to do to the bottom. To get 2 from 6 in the denominator you would have had to divide by 3 so I multiplied 15 by 3 to give make the two fractions equal to each other. If you are not sure if your answers are correct just divide the numerator by the denominator on the calculator and if they are equal values then it's correct.
For a school fund raising project, Aiden and Jacob collected money in the ratio 3.5;6.8. Jacob collected $115.50 more than Aiden. Find the total amount of money they collected
Answers
Answer:
Step-by-step explanation:
what is the lateral surface area of a right circular cylinder with height 10cm and base area 154cmsquare (use pi = 22/7)
Answers
Answer:
Lateral surface area = 154cm²
Step-by-step explanation:
h = 10cm
pi = 22/7
base area = 154cm²
2πrh = 154cm²
2 × 22/7 × r × 10 = 154
440/7 × r = 154
r = 154 ÷ 440/7 = 2.45
r = 2.45
LSA of right circular cylinder = 2πrh
2 × 22/7 × 2.45 × 10
44/7 × 24.5 = 1078/7
1078/7 = 154
Therefore, lateral surface area = 154cm²
Hope it helps:)
Identify the solution for the system of equations graphed here.
A) (1,1)
B) (-1,1)
C) (1,-1)
D) (-1, -1)
Answers
Answer:
c) (1,-1)
Step-by-step explanation:
I hope this helps
Help me out I need it
Answers
Answer:
27 Chairs (See below)
Step-by-step explanation:
A: You will put 9 chairs in each row, and have 3 rows.
For a visual example I will use the symbol "O" for the number of chairs.
OOOOOOOOO
OOOOOOOOO
OOOOOOOOO
B: The multiplication problem, using the information 3 Rows, and 9 Chairs in each row, is 9x3
C: The factors are the numbers you are multiplying with, and the product, is your outcome when you multiply the two factors, 9, and 3.
So your answer is 27.
Hope this helps!
please help..
Evaluate
10 % of 9900
Answers
Answer:
990
Step-by-step explanation:
9900 * 10/100
= 990
Hello there!
First, divide by 100:
99
Now, multiply by 10:
990 is the answer.
Hope it helps!
~Just a determined gal
#CarryOnLearning
[tex]MysteriousNature[/tex]
If an airplane in a 1 in : 2 ft scale drawing is 6 inches long, how long is the actual airplane? 12 ft 24 ft 8 ft 3 ft.
Answers
Answer:
the answer is 12ft hope that helps
In one day a museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adults and how many children were admitted to the museum?
I NEED IN STEP BY STEP SO I CAN UNDERSTAND!!!!
Answers
Answer:
153 adults and 168 children
Step-by-step explanation:
x - number of adults
y - number of children
[tex]\left \{ {{6x + 4y = 1590} \atop {x + y = 321}} \right.[/tex]
[tex]\left \{ {{6x+4y = 1590} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{6*(321 - y)+4y = 1590} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{1926 -6y+4y=1590} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{1926 -2y=1590} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{-2y=1590-1926} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{-2y=-336} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{y=168} \atop {x=321-y}} \right.[/tex]
[tex]\left \{ {{y=168} \atop {x=321-168}} \right.[/tex]
[tex]\left \{ {{y=168} \atop {x=153}} \right.[/tex]
Is it all clear?
Can someone help me please
Answers
Answer:
A
Step-by-step explanation:
4×-12
Answer is letter A
hope it help^^
What is 3.468 rounded to the nearest tenth of an ounce?
Answers
Answer:
3.5
Step-by-step explanation:
The tenths place would be 0.1
So in 3.468 the 4 is in the tenths place
Look at the number to the right of it, if it is greater than 5 you round up and if it is 4 or lower you stay the same.
For 3.468 you would round up
So you would have 3.5
Have A GREAT day man
Sincerly, lipor
[tex]if \: \sqrt{2} + \sqrt{3} \div 3 \sqrt{2} - 2 \sqrt{3} = x + \sqrt{6y} \: find \: the \: value \: of \: x \: and \: y[/tex]
Answers
Given info:- [tex]\textsf{If (√3+√2)/(3√2-2√3) = x+y√6,} [/tex]
[tex]\textsf{then find the value of x and y?}[/tex]
Explanation:-
[tex]\textsf{Consider LHS=}\\[/tex]
[tex] \rm \: \frac{ \sqrt{2 } + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ [/tex]
[tex] \rm \: = \frac{ \sqrt{2 } + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ [/tex]
[tex] = \rm \: \frac{( \sqrt{2 } + \sqrt{3} )(3 \sqrt{2} + 2 \sqrt{3} )}{(3 \sqrt{2} - 2 \sqrt{3})(3 \sqrt{2} + 2 \sqrt{3}) } \\ [/tex]
[tex] \rm \: = \frac{3 \sqrt{2 \times 2} + 2 \sqrt{3 \times 2} + 3 \sqrt{2 \times 3} + 2 \sqrt{3 \times 3} }{(3 \sqrt{2} {)}^{2} - (2 \sqrt{3} {)}^{2} } \\ [/tex]
[tex] \rm = \frac{3 \times 2 + 2 \sqrt{6} + 3 \sqrt{6} + 2 \times 3}{18 - 12} \\ [/tex]
[tex] \rm \: = \frac{12 + 5 \sqrt{6} }{6} \\ [/tex]
[tex] \rm \: = 2 + \frac{5}{6} \sqrt{6} \\ [/tex]
[tex] \rm \therefore \: 2 + \frac{5}{6} \sqrt{6} = x + y \sqrt{6} \\ [/tex]
[tex]\textsf{On, comparing with RHS we notice that:}\\[/tex]
[tex]\textsf{the value of x = 2 and y = 5/6.}\\[/tex]
[tex]\bf{Read more:}[/tex]
[tex]\textsf{Similar questions}\\[/tex]
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