side, s = 1/3ft
area, a = 6s^2
a = 6 × (1/3)^2 = 6/9 = 2/3 sq ft
The temps that we know are colder than -5deg C
first anything below or less than -5degC will be colder in temperature because it has a lower or smaller value, so it is colder overall.
3 examples would be x<-5deg C
x= -6deg C
x= -7deg C
x= -8deg C
so plugging in x value we get -6<-5 -7<-5 -8<-5
If you are needing anything else please just ask! Thank you!
Answer:
7. B 8. C 9. C
Step-by-step explanation:
7. y is greater than because the area above the line is shaded and since the line is dashed there it's not greater than or equal to.
8. C provides the correct y values for all of the x values.
9. The point (6, 3) produces -3 ≥ -8 which is the only one that is correct out of all the points provided.
1) The perimeter is the sum of the lengths of the straight edge (the diameter of the semicircle) and the length of the arc of the semicircle.
The circumference of a 4 ft circle is
π*diameter = π*4 ft ≈ 12.566 ft
The semicircle will have a length that is half that, 6.283 ft. When this length is added to the diameter, the perimeter is found to be
Perimeter = 4 ft + 6.283 ft ≈ 10.3 ft.
2) The area of a circle is given by the formula
A = (π/4)d²
For a diameter of 15 inches, the area is
A = (π/4)(15 in)² = 56.25π in²
A ≈ 176.7146 in²
The area of the circle is about 176.71 in².
When we use arcsine, we are finding the angle while giving the trigonometric ratio.
Arcsin(u) = theta can be rewritten as:
sin(theta) = u
Sine is opposite over hypotenuse, so u/1 means that the side opposite to theta (the y value) is u, and the hypotenuse is 1.
We can use Pythagorean Theorem to find the adjacent (x value).
1^2 - u^2 = x^2
x = sqrt(1-u^2)
Back to the original question, we are trying to find cos(arcsin(u)). We just solved all the sides for our triangle using arcsin(u). Now we need to do cos(u).
Cosine is adjacent over hypotenuse.
So our answer is sqrt(1-u^2)/1
Or just sqrt(1-u^2)
