D (-4; -2)
C (1; 2)
M (x; y)
DM+MC=CD
CM:MD=1:3
M: x= x(C)-(x(C) - x(D))/4=1-(1-(-4))/4=1 - (6/4)=1 - 1,5= - 0,5
M: y=y(C)-(y(C) - y(D))/4=2-(2-(-2))/4=2 - (4/4)=2 - 1=1
M(x; y)=M( -0,5; 1)
<h3>
Answer: 127</h3>
=================================================
Explanation:
The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.
The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.
Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.
So we have these scores
- first exam = 70
- second exam = 80
- third exam = 43
- fourth exam = x+5
Adding up these scores and dividing by 4 will get us the average
(sum of scores)/(number of scores) = average
(70+80+43+x+5)/4 = 80
(x+198)/4 = 80
x+198 = 4*80
x+198 = 320
x = 320 - 198
x = 122
So the student got a score of x+5 = 122+5 = 127 on the fourth exam.
Answer:
Step-by-step explanation:
rectangular prism=lwh
cylinder=πr²h
cone=1/3 πr²h
sphere=4/3 πr³
Answer:

Step-by-step explanation:
∠E = 1/2 (mCD - mAB)
<u>Where mCD = 110° and mAB = 30°</u>
∠E = 1/2 (110° - 30°)
∠E = 1/2 (80°)
∠E = 40°
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>