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iren [92.7K]
3 years ago
12

Have an amazing day :) Will give brainlst! Pls help

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0
not really sure but i need these points
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Evaluate expressions- if A=4 and B=9 what is the value of the following expression
lys-0071 [83]

Answer:

68

Step-by-step explanation:

just put the value of a and b in the expression and you will get the answer

20/a+7.b is

20/4+7×9

5+63

68

7 0
3 years ago
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A certain bamboo plant grows 1.2 feet in one day. At that rate, how many feet could the plant grow in 0.5 day?
Rainbow [258]

Answer:

0.6 feet in half a day

Step-by-step explanation:

If it grows 1.2 feet in 1 full day, then half of that 1.2÷2=0.6

5 0
2 years ago
There are 32 female performers in a dance recital. The ratio of men to women is 3:8. How many men are in the dance recital??????
nydimaria [60]

Answer:

12

Step-by-step explanation:

3/8=x/32

8x=96/32

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6 0
3 years ago
If f(x) = sin(x/2), then there exists a number c in the interval pie/2 < x < 3pie/2 that satisfies the conclusion of the m
elena-14-01-66 [18.8K]
For the answer to the question above, 
The mean value theorem states the if f is a continuous function on an interval [a,b], then there is a c in [a,b] such that: 
<span>f ' (c) = [f(b) - f(a)] / (b - a) </span>
<span>
So [f(a) - f(b)] ( b - a ) = [sin(3pi/4) - sin(pi/4)]/pi </span>
= [sqrt(2)/2 - sqrt(2)/2]/pi = 0 
So for some c in [pi/2, 3pi/2] we must have f ' (c) = 0 

In general f ' (x) = (1/2) cos (x/2) 
We ask ourselves for what values x in [pi/2, 3pi/2] does the above equation equal 0. 
0 = (1/2) cos (x/2) 
0 = cos (x/2) 
x/2 = ..., -5pi/2, -3pi/2, -pi/2, pi/2, 3pi/2, 5pi/2,... 
x = ..., -5pi, -3pi, -pi, pi. 3pi, 5pi, .... 
and x = pi is the only solution in our interval. 

So c = pi is a solution that satisfies the conclusion of the MVT
3 0
3 years ago
WANT FREE 20 POINTS + BRAINLIEST? ANSWER THIS GEOMETRY QUESTION CORRECTLY AND I GOT YOU :)
uysha [10]

Answer:

1.x

2.z

3.v

Step-by-step explanation:

just took the test sorry if i'm wrong

8 0
3 years ago
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