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lara [203]
3 years ago
6

I need help on 5 please help fast

Mathematics
2 answers:
djyliett [7]3 years ago
8 0
 the correct answer is 83.

yyyyyeeeeeeaaaahhhhhh bbbbooooiii
marysya [2.9K]3 years ago
6 0
The equation you would use is 
mean= Test1+test2+test3+test4+test5+Test6+x/Total number of tests

so 83=89+78+69+81+89+90+X/7
First mutiple both sides by 7 so 
581=496+X
you then substract the 496 and get 85
so she would need a 85 on the 7th test
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Please answer this and thank you<br><br> If DE=4x+10, EF=2-1, and DF=9x-15, find DF.
professor190 [17]
I hope that 2-1 means 21.

DE + EF = DF
4x + 10 + 21 = 9x - 15
4x + 31 = 9x - 15
31 = 9x - 4x - 15
31 = 5x - 15
46 = 5x
x = 9 1/5 = 9.2

You want 9x - 15
9*9.2 - 15
82.8 - 15
67.8 <<<<< answer. 
If I have given the wrong meaning to 2-1 (which could mean 1) leave another note and I will correct my work. 
8 0
3 years ago
Histogram help!(look at the picture)The histogram below shows the admission cost of some theme parks across the country.How many
Andrei [34K]

Answer:

12

Step-by-step explanation:

5+7=12

5 have 7-14

7 have 15-22

The final answer is 12

3 0
3 years ago
Example:
Aleonysh [2.5K]

Answer:

AB // CD and AD // BC, then ABCD is a parallelogram

Step-by-step explanation:

Step 2: Show AD // BC

A = (-3 , 3) and D = (0 , 0)

Find the slope of AD

m_{AD}=\frac{0-3}{0--3}=\frac{-3}{3}=-1

B = (2 , 5) and C = (5 , 2)

m_{BC}=\frac{2-5}{5-2}=\frac{-3}{3}=-1

The slope of AD = the slope of BC

So AD // BC because they have the same slope

Step 3:

Each two opposite sides are parallel in quadrilateral ABCD

So the quadrilateral is a parallelogram

8 0
3 years ago
I need help it's so confusing!​
jasenka [17]
Try looking up the answer
5 0
3 years ago
For what values of t on the interval [0, 21 ]is the instantaneous velocity positive​ (the projectile moves​ upward)?
OLEGan [10]

Answer:

0 <=t<=21

Step-by-step explanation:

Projectile is Moving upwards on an interval of (0 to 21), if we plot Velocity vs Time and denote positive y-axis above 0 and negative y-axis below  0(for velocity), then from 0 to 21 t projectile is moving upwards and has positive velocity, when the projectile reaches the top of it's motion and returns back down to ground it's velocity is negative and is plotted below the y =0 (note that is for t > 21).

hence for the interval 0 <=t <=21 the instantaneous velocity is positive (Note, instantaneous velocity is also the derivative of the velocity or the slope ).

8 0
3 years ago
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