<span>y + 4 ≤ 0
----------------------
Subtract 4 from each side
</span><span>y + 4 - 4 ≤ 0 - 4
</span>y ≤ -4
y ≤ -4 is the answer
Answer:
18 children weighed less than 50 pounds.
Step-by-step explanation:
Ok, adding into the explanation what the heck this kind of a table is. The "stem" column is the tens place of each of the children's weights. The "leaf" is the ones digit of each of the children's weights. So in the row where the stem is 2, there were 2 kids weighing 23 and 25 pounds.
For your question, we have to add up how many students have a weight with a "stem" less than 5, because that's where 50 pounds starts.
8 + 2 + 5 + 3 = 18 children weighed less than 50 pounds.
0.3m=$4.41
just divide $4.41/0.3= m
m=14.7
Answer:
The area of APC is 70m². The area of triangle PMC is 35m².
Step-by-step explanation:
Let the area of triangle ABC be x.
It is given that AM is median, it means AM divides the area of triangle in two equal parts.
.....(1)
The point P is the midpoint of AB, therefore the area of APC and BPC are equal.
......(2)
The point P is midpoint of AB therefore the line PM divide the area of triangle ABM in two equal parts. The area of triangle APM and BPM are equal.
.....(3)
The area of triangle APM is 35m².
![\text{Area of }\triangle APM=\frac{x}{4}](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%5Ctriangle%20APM%3D%5Cfrac%7Bx%7D%7B4%7D)
![35=\frac{x}{4}](https://tex.z-dn.net/?f=35%3D%5Cfrac%7Bx%7D%7B4%7D)
![x=140](https://tex.z-dn.net/?f=x%3D140)
Therefore the area of triangle ABC is 140m².
Using equation (2).
![\text{Area of }\triangle APC=\frac{x}{2}](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%5Ctriangle%20APC%3D%5Cfrac%7Bx%7D%7B2%7D)
![\text{Area of }\triangle APC=\frac{140}{2}](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%5Ctriangle%20APC%3D%5Cfrac%7B140%7D%7B2%7D)
![\text{Area of }\triangle APC=70](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%5Ctriangle%20APC%3D70)
Therefore the area of triangle APC is 70m².
Using equation (3), we can say that the area of triangle BPM is 35m² and by using equation (2), we can say that the area of triangle BPC is 70m².
![\triangle BPC=\triangle BPM+\triangle PMC](https://tex.z-dn.net/?f=%5Ctriangle%20BPC%3D%5Ctriangle%20BPM%2B%5Ctriangle%20PMC)
![70=35+\triangle PMC](https://tex.z-dn.net/?f=70%3D35%2B%5Ctriangle%20PMC)
![35=\triangle PMC](https://tex.z-dn.net/?f=35%3D%5Ctriangle%20PMC)
Therefore the area of triangle PMC is 35m².
Answer:
(#14 - supplementary) x = 10.5
(#15 - supplementary) x = 10
Explanation:
For these problems, we must know that a supplementary angle can be viewed as a sum of angles that add to 180 degrees. With this known, we can write our equations to find x.
(#14)
15x - 12 + 5x - 18 = 180
20x - 30 = 180
20x = 210
x = 10.5
(#15)
6x + 13 + 14x - 33 = 180
20x - 20 = 180
20x = 200
x = 10
Hope this helps.
Cheers.