In the figure below, LINE BP bisects angle ABC. Line CP bisects angle ACB. Angle ACP is 30 degrees. Angle ABP is 38 degrees. Name the measures of two other angles.
Describe how you found the measures.
Answers
Answer:
Angle ACB is 60 degrees and angle ABC 76 degrees
Step-by-step explanation:
If an angle is bisected it means it is cut in half. If half angle ACB is 30 degrees than the entire angle is 30 x 2 or 60 degrees. I did the same thing for angle ABC where its whole is 38 x 2 or 76 degrees.
Related Questions
Mary covered her kitchen floor with 10 tiles. The floor measures 6 feet long by 5 feet wide, The tiles are each 3 feet long and w feet wide. Write an equation to represent the situation.
What is 4/15 divide 10/13
How are the values of the two 4s related in .044
Prime numbers between 30 and 40
b = w + 11
b = 3w − 15
Which of the following methods is correct to find Betsey's and Waldo's age?
Solve w + 11 = 3w − 15 to find the value of w.
Solve b + 11 = 3b − 15 to find the value of b.
Write the points where the graphs of the equations intersect the x-axis.
Write the points where the graphs of the equations intersect the y-axis.
Answers
Answer:
Option 1st is correct
Solve w + 11 = 3w − 15 to find the value of w.
Step-by-step explanation:
Here b represents the Betsey's age and w represents the Waldo's age
As per the statement:
Betsey is 11 years older than Waldo. Betsey's age is 15 years less than three times Waldo's age.
The system below models the relationship between b and w:
Equate these equations to solve for w, we have;
Add 15 to both sides we have;
Subtract w from both sides we have;
Divide both sides by 2 we have;
13 = w
or
w = 13 years
Substitute in b = w+11 we have;
b = 13 +11 = 24 years.
b = 24 years
Therefore. following methods is correct to find Betsey's and Waldo's age is:
Answers
I looked at this just before going to bed, and it ruined my night, worrying
about it. Then, this morning, when I actually set pencil to paper, it turned
out not to be such a vicious limbic twister.
You DO need to know how to take a chain derivative, though. I must say,
I'm surprised to see this much Calculus at the middle school level. Maybe
it's MY problem, but I just can't see a way to do it without this derivative,
so I'll just show you how I did it:
The ants begin together, and crawl out along the legs of a right triangle.
-- The first one moves at 4 ft/min, so after 't' minutes, he's 4t-ft from
the vertex. The question concerns what happens when this one has
covered 12-ft. That will happen in (12/4) = 3 minutes after he starts out.
-- The second one starts out 2-min later, and moves at 5-ft/min. So,
after 't' minutes, he's 5(t-2) ft from the vertex.
-- The distance between them is the hypotenuse of the right triangle.
The square of the distance is the sum of the squares of the legs:
D² = (4t)² + [ 5(t-2) ]²
We're actually going to need to know how far apart they are after 3 minutes,
so let's do that now.
D² = (4 x 3)² + (5 x 1)² = (12)² + (5)² = 144 + 25 = 169
D = √169 = 13-ft apart, after the first guy has covered 12 feet.
We will need this shortly.
OK. Now, let's simplify the equation for the distance between them,
(as we get ready to differentiate it).
D² = (4t)² + [ 5(t-2) ]²
D² = 16t² + (25) (t² - 4t + 4)
D² = 16t² + 25t² - 100t + 100
D² = 41t² - 100t + 100
The question asks "How fast is the distance between them changing . . . ? ",
and we know that this is the derivative of 'D' with respect to 't' . It's easy to
take the derivative of the right side of the equation, but what about the derivative
of ' D² ' ? That's the part that I'm not so sure about in middle school, so I'll just
give it to you:
(Derivative of D²) is [ (2D) times (derivative of D) ] .
Now we can take the derivative of each side of the equation:
(2D) x (derivative of D) = 82 t - 100 .
derivative of D = (82 t - 100) / (2D)
That's how fast the distance between them is changing at any time 't' .
After 3 minutes, we calculated 'D' earlier. It's 13-ft
derivative of D = [ (82 x 3) - 100 ] / (2 x 13)
= 146 / 26
= 5.615 feet per second.
Are you confident ? Do you trust me?
I'm not confident at all, and I don't trust myself.
The best I can suggest is for you to go through this and look for my mistakes,
or better yet, sit down with your teacher and go through it for mistakes.
That'll help you think about it and understand it, even if my work is wrong.
What you have to know is that:
Also:
t=time snapshot (in minutes)
x=distance first ant has travelled
y=distance second ant has travelled (perpendicular to the first ant)
D=distance between both ants
--------------------------------------
Now, when:
t=0, x=0, y=0, D=0
t=1, x=4, y=0, D=4
t=2, x=8, y=0, D=8
t=3, x=12, y=5, D=13
Now we quantize time and find the tangent line (slope) between two points on the distance vs time graph at t=3 and t=3.000001, knowing that distance is represented by feet and time is represented in minutes.
When t=3.000001, x=12.000004, y=5.000005 D=13.00000562...
Now we find the slope on the distance vs time graph between the points (3,13) and (3.000001, 13.00000562).
So, at the point (3,13) on the distance vs time graph, the two ants are moving away from each other at approximately 5.62 feet per minute.
Notations you may require to understand problem in the attachment below.
Answers
Answers
Answers
6p-4= 6×5-4 = 30-4= 26
30-4=26
26 is your answer
Answers
Answer:
192
Step-by-step explanation:
Its nothing complicated just add their scores together.
Answer:
331
Step-by-step explanation:
139 + 53 = 192
C = 139
B = 192
139 + 192 = 331