Answer : acute
Right triangle : a^2+b^2=c^2
Acute triangle : a^2+b^2 > c^2
Obtuse triangle:a^2+b^2 < c^2
6^2+7^2 > 9^2
Answer:
To find the x-intercepts we can write:
0 = -1/2(x + 3)² + 4
0 = -1/2(x² + 6x + 9) + 4
0 = -1/2x² - 3x - 9/2 + 4
0 = -1/2x² - 3x - 1/2
0 = x² + 6x + 1
x = -3 ± 2√2 (use quadratic formula)
As a decimal this can be written as x = -5.83, x = -0.17.
Answer:
4:50
Step-by-step explanation:
Thinks it b because it going to be a different size
In order to solve for this question, let's assign a couple variables.
The variable 't' will represent the number of two-point questions, and the variable 's' will represent the number of six point questions.
From the given, we can already form two equations:
t + s = 36
("An exam... contains 36 questions")
2t + 6s = 148
("An exam worth 148 points... Some questions are worth 2 points, and the others are worth 6 points")
Before we begin calculating anything, we can simplify the second equation we made, since all the numbers are divisible by 2:
2t + 6s = 148
t + 3s = 74
Now let's refer back to the first equation. We can subtract both sides by 's' (you could also subtract both sides by 't', but I personally think that this will make solving the equation less difficult):
t + s = 36
t = 36 - s
This effectively gives us a value for the variable 't'. We can assign this value back into our first equation:
t + 3s = 74
(36 - s) + 3s = 74
36 + 2s = 74
2s = 38
s = 19
Input 's' into our second equation to solve for 't':
t = 36 - s
t = 36 - 19
t = 17
There are 17 two-point questions and 19 six-point questions
- T.B.
