Answer:
357.1573 (If you don't round)
Step-by-step explanation:
To find area of a rectangle is to multiply length by width, which in this situation would be 15 by 16.6, which is equal to 249. Next, we need to add this to the area of the semicircle. The area of a semicircle is equal to 1/2 * pi * radius^2 (Because a semicircle is half a circle and a circle's area is pi * radius^2). The radius is equal to half the diameter, which in this case is 16.6. 16.6/2 = 8.3, so the radius is 8.3. If we plug this into the equation, we get that 108.1573 is the area of the circle (If you use 3.14 as pi). 108.1573 + 249 is equal to 357.1573.
The answer would be C because it’s the only fraction that does not equal 0.4
Let y be the unknown number
The other number is x - 12
Answer: It is not possible that two triangles that are similar and not congruent in spherical geometry.
Step-by-step explanation:
For instance, taking a circle on the sphere whose diameter is equal to the diameter of the sphere and inside is an equilateral triangle, because the sphere is perfect, if we draw a circle (longitudinal or latitudinal lines) to form a circle encompassing an equally shaped triangle at different points of the sphere will definately yield equal size.
in other words, triangles formed in a sphere must be congruent and also similar meaning having the same shape and must definately have the same size.
Therefore, it is not possible for two triangles in a sphere that are similar but not congruent.
Two triangles in sphere that are similar must be congruent.
Answer:
(You only need to give one solution)
Step-by-step explanation:
We have the following equation
First, we need to foil out the parenthesis
Now we can combine the like terms
Now, we need to factor this equation.
To factor this, we need to find a set of numbers that add together to get -3 and multiply to give us -4.
The pair of numbers that would do this would be 1 and -4.
This means that our factored form would be
As the first binomial is a difference of squares, it can be factored futher into
Now, we can get our solutions.
The first binomial will produce two complex (Not real) solutions.
So our solutions to this equation are
