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kirza4 [7]
3 years ago
11

Marquise has 200 meters of fencing to build a rectangular garden.

Mathematics
2 answers:
Levart [38]3 years ago
8 0

Answer:

50

Step-by-step explanation:

VikaD [51]3 years ago
5 0

Answer:

2500

Step-by-step explanation:

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Estimate the product by rounding 2 x 6,254
Pavlova-9 [17]
5 × 6,254= 31,300 is the awnser :)
6 0
4 years ago
What is the domain and range of the graph?
Andreas93 [3]

Answer:

Domain= how many boxes there are from each circle to the other, and the range is basically the height. In this case, Domain= 8, Range=2

8 0
3 years ago
2(4x – 12) + 10 – 6x = 12
SashulF [63]

Answer:

idk what the improper fraction would be but

8x-24+10-6x=12

2x-14=12

2x=26

x=13

Step-by-step explanation:

5 0
3 years ago
2^sinx+2^cosx find the minimum value of this trignometry
Anit [1.1K]
Take the deritiivive
ln(2)cos(x)2^{sin(x)}-ln(2)sin(x)2^{cos(x)}
it is zero at pi/4 and it repeats at every pi
minimum at where slope goesfrom negative to positive
so at 5pi/4 and 13pi/4 so at every 2pi interval


evaluate origial at 5pi/4
2^{sin(5pi/4)}+2^{cos(5pi/4)}=
(2^ \frac{- \sqrt{2} }{2} )+(2^ \frac{- \sqrt{2} }{2} )=
2(2^ \frac{- \sqrt{2} }{2} ) or aprox 1.22509
5 0
4 years ago
The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon
Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by x=3.24t-4.20t^2+1.07t^3 where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

x(t)=3.24t-4.20t^2+1.07t^3

x(1)=3.24(1)-4.20(1)^2+1.07(1)^3

x(1)=3.24-4.20+1.07

x(1)=0.11

b) At t= 2 s

x(t)=3.24t-4.20t^2+1.07t^3

x(2)=3.24(2)-4.20(2)^2+1.07(2)^3

x(2)=6.48-16.8+8.56

x(2)=-1.76

c) At t= 3 s

x(t)=3.24t-4.20t^2+1.07t^3

x(3)=3.24(3)-4.20(3)^2+1.07(3)^3

x(3)=9.72-37.8+28.89

x(3)=0.81

d) At t= 4 s

x(t)=3.24t-4.20t^2+1.07t^3

x(4)=3.24(4)-4.20(4)^2+1.07(4)^3

x(4)=12.96-67.2+68.48

x(4)=14.24

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

\triangle x=x(4)-x(0)

\triangle x=14.24-0

\triangle x=14.24

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity  is given by,

\triangle x=x(4)-x(2)

\triangle x=14.24-(-1.76)

\triangle x=14.24+1.76

\triangle x=16

4 0
3 years ago
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