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Mademuasel [1]
3 years ago
5

Calculate the median of 5,10,12,4,6,11,13,15

Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0
The median would be 10.5
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A water well drilling rig has dug to a height of -60 feet after one full day of continuous use.
RideAnS [48]
The answer should be -37.5 feet unless it is 15 more hours then it is -97.5
4 0
3 years ago
the slope of line PQ is m= 1/4. find the equation of the line that is parallel to line PQ and that has the y intercept b= -3
nevsk [136]
For this case we have that the generic equation of the line is given by:
 y = mx + b

 Where,
 m: slope of the line
 b: intersection with the y axis.
 Since the line is parallel to PQ, then the slopes are equal.
 We have then:
 m = 1/4.

 On the other hand we have:
 b = -3

 Substituting values we have:
 y = (1/4) x - 3
 Answer:
 The equation of the line that is parallel to line PQ and that has the and intercept b = -3 is:
 y = (1/4) x - 3
4 0
3 years ago
(6th grade math)
Montano1993 [528]

Answer:

He drank 15.93 ounces.

4 0
3 years ago
Read 2 more answers
1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t
natta225 [31]

The area of the surface given by \vec r_1(u,v) is 1. In terms of a surface integral, we have

1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

By multiplying each component in \vec r_1 by 5, we have

\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}

and the same goes for the derivative with respect to v. Then the area of the surface given by \vec r_2(u,v) is

\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}

8 0
3 years ago
What is the area of a 14m by 5m rectangle?
Lisa [10]
Area=L multiply by W
14×5=70m
7 0
3 years ago
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