Answer: Conjecture: There is no triangle with side lengths N, 2N, and 3N (where N is a positive real number)
Proof:
We prove this by contradiction: Suppose there was an N for which we can construct a triangle with side lengths N, 2N, and 3N. We then apply the triangle inequalities tests. It must hold that:
N + 2N > 3N
3N > 3N
3 > 3
which is False, for any value of N. This means that the original choice of N is not possible. Since the inequality is False for any value of N, there cannot be any triangle with the given side lengths, thus proving our conjecture.
<h3>
Answer: Lower right corner (ie southeast corner)</h3>
In this graph, it is impossible to draw a single straight vertical line through more than one point on the yellow line. Therefore, we conclude that this graph passes the vertical line test, which indicates we have a function.
In contrast, the upper left corner fails the vertical line test. Note the left-most pair of points are vertically stacked together. A single vertical line goes through these two points. So this is one possible way to show the graph does not pass the vertical line test, and thereby making this not a function. The upper right corner and lower left corner has the same idea as the upper left corner, so they aren't functions either.
Answer:
z≥−3
Step-by-step explanation:
1 Subtract 88 from both sides.
5z≥−7−8
2 Simplify -7-8−7−8 to -15−15.
5z≥−15
3 Divide both sides by 55.
z≥− 15/5
4 Simplify 15/5 to 3
z≥−3
Answer:
A. -1
General Formulas and Concepts:
<u>Algebra I</u>
Slope-Intercept Form: y = mx + b
Step-by-step explanation:
<u>Step 1: Define</u>
y = -2x - 1
<u>Step 2: Break function</u>
Slope <em>m</em> = -2
y-intercept <em>b</em> = -1
Step-by-step explanation:
f(x) = x² + x + 3/4
in general, such a quadratic function is defined as
f(x) = a×x² + b×x + c
the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 1
c = 3/4
x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =
= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =
= (-1 ± sqrt(2)i)/2
x1 = (-1 + sqrt(2)i) / 2
x2 = (-1 - sqrt(2)i) / 2
remember, i = sqrt(-1)
f(x) has no 0 results for x = real numbers.
for the solution we need to use imaginary numbers.