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Arisa [49]
3 years ago
11

Can I get help and no links pls

Mathematics
2 answers:
kodGreya [7K]3 years ago
5 0
The answer is 30, can I get the brainliest answer
Svet_ta [14]3 years ago
3 0
The person was correct it is 3
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A radar gun was used to record the speed of a swimmer (in meters per second) during selected times in the first 2 seconds of a r
Natalka [10]
Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]

the area of each trapezoid is (v(t1)+v(t2))/2 times width

for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8

2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24

3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75

4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7

add them all up
0.8+2.24+1.75+0.7=5.49

5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters
8 0
3 years ago
Ms. Cannon is a tutor. She charges $ 25 for travel time and $ 35 for each hour, h, she tutors. What expression represents the to
wolverine [178]

Answer:

25 + 35h

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The sum of two numbers is 58. The larger number is 6 more than the smaller number. What are the numbers?
kenny6666 [7]

Answer:

Numbers are 32, and 26.

Step-by-step explanation:

58 / 2 = 29

29 + 3 = 32 (large number)

29 - 3 = 26 (small number)

8 0
3 years ago
Read 2 more answers
Two hundred students are going to dine with the principal as their reward for perfect attendance. A random survey of 25 of these
Karolina [17]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
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How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x​
ivanzaharov [21]

\Large \mathbb{SOLUTION:}

\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes}  \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}

5 0
3 years ago
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