Answer:the system has no solutions
Step-by-step explanation:
Answer:
We need more info.
Step-by-step explanation:
<h3>Answer: True</h3>
Explanation:
Imagine that f(x) was a road without any gaps. The road can be straight or it can be curved somehow.
Now imagine that a horizontal line is the border between two countries. The country up north is the positive country and the southern country is negative.
f(1) > 0 means we're in the northern country since f(x) is positive here
then f(3) < 0 means we're now in the southern negative country
Somewhere along the road we must have crossed the border at least one time. This must be the case because the road does not have any gaps in it and we cannot teleport. This is what it means to be a continuous function. To draw a continuous curve, you cannot lift your pencil up.
So because f(x) transitions from positive to negative, this means f(x) = 0 at some point at least once. So that's why there exists a c such that f(c) = 0
First you must state that the relationship is proportional. It is because it goes into the origin.
y = 2.5x
k = 2.5 And 2.5 is the constant of proportionality.
Answer:
Not really
Step-by-step explanation:
NOT NECESSARILY would a triangle be equilateral if one of its angles is 60 degrees. To be an equilateral triangle (a triangle in which all 3 sides have the same length), all 3 angles of the triangle would have to be 60°-angles; however, the triangle could be a 30°-60°-90° right triangle in which the side opposite the 30 degree angle is one-half as long as the hypotenuse, and the length of the side opposite the 60 degree angle is √3/2 as long as the hypotenuse. Another of possibly many examples would be a triangle with angles of 60°, 40°, and 80° which has opposite sides of lengths 2, 1.4845 (rounded to 4 decimal places), and 2.2743 (rounded to 4 decimal places), respectively, the last two of which were determined by using the Law of Sines: "In any triangle ABC, having sides of length a, b, and c, the following relationships are true: a/sin A = b/sin B = c/sin C."¹