#8
24 • 24= 576 and 25 • 25= 625
76 • 76= 5,776 and 77 • 77= 5, 929
242 • 242= 58, 564 and 243 • 243= 59, 049√
766 • 766= 586, 756 and 767 • 767=588, 298
( by the way,
I'm squaring it because it's a square.)
#9
3x -5≥ 5x-25
-3x. -3x
-----------------
---> -5 ≥2x -25
+25. +25
---------------------
---->20≥2x
20/2 ≥ 2x/2
----->10≥x, the greatest 'x' can be is 10.
Answer:
BC = 54
Step-by-step explanation:
Don't get caught on this question. There really is enough information.
The vertical line through D can be shown to be a perpendicular bisector of AC.
That means AB = BC
6x = 9x - 27 Add 27 to both sides
6x + 27 = 9x Subtract 6x from both sides
27 = 9x - 6x
27 = 3x Divide by 3
27/3 = x
x = 9
There are two ways to find BC.
1) Find AB
AB = 6 * 9 = 54
2) Find BC
BC = 9x - 27
BC = 9*9 - 27
BC = 81 - 27
BC = 54
Good thing to remember for a test. Remember that BD is the perpendicular bisector of AC. Both segments making up AC are equal.
Answer:
The absolute value function shows up in the world around us in many different areas. Suppose you are driving down the road and you look out your window and see a speed limit sign that says the speed limit is 50mph. You look at your speedometer and see that you're driving at 45mph, so you are going 5mph below the speed limit. Notice that even though you are going 5mph below the speed limit, we don't say you are going -5mph from the speed limit. We just state the difference from 50mph as a positive value. A road sign shows a vehicle's speed as the vehicle passes. SPEED LIMIT 30 YOUR SPEED Part A: The sign blinks for vehicles traveling within 5 mi/h of the speed limit. Write and solve an absolute value inequality to find the minimum and maximum speeds of an oncoming vehicle that will cause the sign to blink. Part B: Another sign blinks when it detects a vehicle traveling within 2 mi/h of a 35 mi/h speed limit. Write and solve an absolute value inequality to represent the speeds of the vehicles that cause the sign to blink.
Step-by-step explanation:
