124° Is your answer.
The triangle Is an isosceles triangle which means the two bases will be the same angle. 180° is this sum of all angles in a triangle. If 62 is the right base angle then 62 must be the left base angle since is an isosceles triangle. Add 62 and 62 together then subtract from 180. That would give you 54° as the third angle angle at top. Since you are looking for a Z, we now know that z and 54° are symmetry and equal 180 when added together. We then subtract 54 from 180 to get the angle of z.
Answer:
Step-by-step explanation:
Answer:
15x² - 3x - 7
Step-by-step explanation:
(12x² + 2x) - (-3x²+ 5x + 7)
First open the parentheses by applying the distributive property.
To do this, multiply each term you have inside (-3x⅔ + 5x + 7) by -1.
Thus, you would have the following:
12x² + 2x + 3x² - 5x - 7
Add like terms
12x² + 3x² + 2x - 5x - 7
15x² - 3x - 7
Answer:
The slope is y=0.59x so 0.59
The y intercept is 0
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
