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Elena L [17]
3 years ago
11

If two equilateral triangles have equal perimeters, they must also have equal areas? True False

Mathematics
2 answers:
Nikitich [7]3 years ago
5 0
That is true.  Since the area of a triangle is 1/2bh the b and h don't change if the triangles are the same
FinnZ [79.3K]3 years ago
3 0
The statement is true.

Hope it helps!
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Find the missing angle
dangina [55]

Answer:

95 degrees

Step-by-step explanation:

The given shape is a triangle.

Let the missing angle be " x ".

Sum of interior angles in a triangle is 180 degrees,

x + 45 + 40 = 180

x + 85 = 180

Subtract 85 on both sides,

x = 180 - 85

x = 95 degrees

6 0
2 years ago
Read 2 more answers
A basketball player is 7 ft tall. He shoots the ball into the basket and the path of the
emmainna [20.7K]

Answer:

52

Step-by-step explanation:

Given the the equation h(t) =-15t^2+ 30t+7 where t is the time in second

So to find the ball's maximum height we can apply the vertex formula:

t= \frac{-b}{2a}  

to find the "x" value of the vertex, then plug that value into the original equation as a substitute for "x".

Standard quadratic form is: ax^2+bx+c

=> a=15, b=30 in our given equation

<=> t = \frac{-30}{2*-15} =1

When t =-1 we have h(t) = 15*1^2+ 30*1+7 = 52

So the ball's maximum height is:  52

Hope it will find you well.

6 0
2 years ago
Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 and the length of B C is 6. Angle A B
lawyer [7]

Answer:

B) √3

Step-by-step explanation:

here is the picture

5 0
2 years ago
Consider the following set of processes, with the length of the CPU burst given in milliseconds:
Degger [83]

Answer:

a) Drawing for grant charts that illustrates execution of the process is in the image attached

b) To find turn around time we use: completion time - arrival time

•Turn around time for FCFS scheduling algorithm will be:

P1 = 2-0= 2

P2= 3-0 = 3

P3 = 11-0 = 11

P4 = 15-0 = 15

P5 = 20-0 = 20

• Turn around time for SJF scheduling algorithm

P1 = 3-0= 3

P2 = 1-0 = 1

P3= 20-0 = 20

P4= 7-0 = 7

P5 = 12-0 = 12

• Turnaround time for non-preemptive algorithm

P1 = 15-0 = 15

P2 = 20-0 = 20

P3 = 8-0 = 8

P4 = 19-0 = 19

P5 = 13 - 0 = 13

• Turnaround time for RR

P1 = 2-0=2

P2= 3-0= 3

P3= 20 - 0 = 20

P4 = 19-0 = 19

P5 = 18-0 = 18

c) To find waiting time we use (turnaround time - burst time)

•Waiting time for FCFS

P1= 2-2 = 0

P2 = 3-1 = 2

P3 = 11-8 = 3

P4 = 15-4 = 11

P5 = 20-5= 15

• Waiting time for SJF

P1= 3-2 = 1

P2 = 1-1 = 0

P3 = 20-8 = 12

P4 = 7-4 = 3

P5 = 12-5 = 7

• Waiting time for non-preemptive

P1= 15-2 = 13

P2 = 20-1 = 19

P3 = 8-8 = 0

P4 = 19-4 = 15

P5 = 13 - 5 = 8

• Waiting time for RR

P1 = 2-2 = 0

P2 = 3-1 = 2

P3 = 20-8 =12

P4 = 13-4= 9

P5= 18-5=13

d) Average waiting time

•'For FCFS

= (0+2+3+11+15)/5

= 31/5

= 6.2milliseconds

•average waiting time For SJF

(1+0+12+3+7)/5

=23/5

= 4.6milliseconds

• Average waiting time For non-preemptive

(13+9+0+15+8)/5

=55/5

=11milliseconds

• average waiting time For RR

(0+2+12+9+13)/5

=7.2milliseconds

Since the SJF algorithm has 4.6milliseconds, it results in the minimum average waiting time.

5 0
3 years ago
Which item would have a mass of about 1 gram?   A. dog   B. thumb tack   C. notebook   D. house
Ierofanga [76]
The answer is B. thumb tack.
5 0
3 years ago
Read 2 more answers
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