Answer:
y=69
Step-by-step explanation:
first step:when y=26 and x=21
c=y-x
c=26-21
c=5
second step: when y is unknown and x=64
To find y we simply add the value of c to x
y=x+c
y=64+5
y=69
Given:
Right triangles:
To find:
The value of c and value of d
Solution:
<u>In the first right triangle:</u>
θ = 45°
Opposite side to θ = 5
Hypotenuse = c
The value of sin 45° =
Do cross multiplication, we get
<u>In the second right triangle:</u>
θ = 60°
Opposite side to θ = 4
Adjacent side to θ = d
Do cross multiplication, we get
The value of and .
Answer:
Q₂ = 10
Step-by-step explanation:
Given data set: 3, 9, 12, 4, 6, 40, 25, 14, 10
Q₂ is the median (the middle value) of the <u>given data set</u> when the data set is <u>ordered</u> from <u>smallest to largest</u>.
<u>Order the values</u> in the data set:
3, 4, 6, 9, <u>10</u>, 12, 14, 25, 40
As the number of values is <u>odd</u>, the middle value is the 5th value.
Therefore, the median (Q₂) of the given data set is 10.
If the number of data values is <u>even</u>, and therefore there are <u>two middle values</u>, the median is <u>halfway</u> between them.
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum). If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2)) =0.0962250K (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
B
0/05×75=3/75
0/03×450=13/5
13/5+3/75=17/25