Answer:
Step-by-step explanation:
A)
the sum of any two sides of a right triangle more than any other side
14+15>29, x is less than 29
14+x>15
x>1
15+x>14
x>-1
1<x<29
lets use Pythagorean Theorem
14^(2)+15^(2)=c^2
196+225=c^2
421=c^2
plusminus sqrt(421)=c
distance can't be negative, so:
c=sqrt(421)
c=20.5182845287
sqrt(421) is the largest possible right triangle side
1<x
sqrt(421)
B) as we can see from above, the largest possible length of the third side is 14/sqrt(421) or about 20.5182845287ft
C)sqrt(421) by 14 by 15
tan(x1)=15/14
x1=tan^-1(15/14)
x1=47 degrees
tan(x2)=14/15
x2=tan^-1(14/15)
x2=43 degrees
You would have to bring it closer to the stage. At that point initially in the diagram, the angle is 25°. As you can see, the angle is small. But when you bring it closer to the stage, the opening would also increase. Otherwise, if you move it farther away, the opening becomes even smaller.
70 m * (100cm)/(1m) * (1 in)/(2.54 cm) = 2755.9 in
.
<span>Rough check, 1 m = 39.36 in approx , 70*39.36 = 2755.2
</span>
I think that's how it's done. As far as explaining it goes, I can't do that. I had to ask my math teacher and she wrote the calculations down.
Answer:
You can either do a back to back stem and leaf plot, where you would have double the values. In a normal stem and leaf plot you would just have one set of 3's where you would put all the values that start with 3 in that column. A back to back is the same but instead you would have two 3 values, where anything that is higher than 5 would be in the second value of 3, but anything lower would be in that first value of 3.
Answer:
6 classes
Step-by-step explanation:
Let n represent the number of classes. Then the total study time is ...
(2/3)n = 4 . . . . . hours
n = (3/2)(4) = 6 . . . . . . . multiply by the inverse of the n coefficient
Sung Lee has 6 classes.
