his score was increased... so if he is now at 67 and it was increased by 7 so his first grade would have been 60.
Answer:
m∠ABE = 27°
Step-by-step explanation:
* Lets look to the figure to solve the problem
- AC is a line
- Ray BF intersects the line AC at B
- Ray BF ⊥ line AC
∴ ∠ABF and ∠CBF are right angles
∴ m∠ABF = m∠CBF = 90°
- Rays BE and BD intersect the line AC at B
∵ m∠ABE = m∠DBE ⇒ have same symbol on the figure
∴ BE is the bisector of angle ABD
∵ m∠EBF = 117°
∵ m∠EBF = m∠ABE + m∠ABF
∵ m∠ABF = 90°
∴ 117° = m∠ABE + 90°
- Subtract 90 from both sides
∴ m∠ABE = 27°
Answer:
-16 > n or n < -16
Step-by-step explanation:
First, we can write the inequality and distribute.
-6n-12 > 3(-n+12)
-6n-12 > -3n+36
Get rid of a variable first (it makes solving a lot easier). We can add 6n to both sides.
-6n-12 > -3n+36
<u>+6n +6n</u>
-12 > 3n+36
Now, we can subtract 36 on both sides.
-12 > 3n+36
<u>-36 -36</u>
-48 > 3n
Divide by 3 on both sides.
-48 > 3n
<u>/3 /3</u>
-16 > n or (if it's easier) n < -16
Hope this helps!! Have a wonderful day C:
Answer:
r < -12 and r > -6
Step-by-step explanation:
The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at
(-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
