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Lesechka [4]
3 years ago
13

Marisol rode her bicycle each day for five days. Estimate how far she biked in all. Round each number to the nearest whole numbe

r. monday 12.3 tuesday 14.1 wednesday 17.7 thursday 11.8 friday 15.2
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
4 0

Answer:

The answer is 71 miles

Step-by-step explanation:

According to the decimal rule, the Monday's 12.3 will be rounded off to 12,Tuesday 14.1 will be rounded off to 14, Wednesday 17.7 will be rounded off to 18 as .7 is greater than .5,Thursday 11.8 is rounded off to 12 and Friday to 15

To calculate how far she biked in all in five days we add all the distances she covered every day.So,

12 miles+14 miles+18 miles+12 miles+15 miles= 71 miles.    

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Laurague here is your answer:)
gayaneshka [121]
Got it :)Thank you!!!!!!!
3 0
3 years ago
What is the 4th term of the linear sequence 27,25,23,21,19
sashaice [31]

Answer:

I believe it is 17

Step-by-step explanation:

subtract 2

5 0
3 years ago
Find out the number of combinations and the number of permutations for 8 objects taken 6 at a time. Express your answer in exact
umka2103 [35]

Solution:

The permutation formula is expressed as

\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\  \end{gathered}

The combination formula is expressed as

\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\  \\  \end{gathered}

where

\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}

Given that 6 objects are taken at a time from 8, this implies that

\begin{gathered} n=8 \\ r=6 \end{gathered}

Thus,

Number of permuations:

\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}

Number of combinations:

\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}

Hence, there are 28 combinations and 20160 permutations.

7 0
1 year ago
The graph shows the functions f(x), p(x), and g(x):
Zinaida [17]
A) The solutions to this set of equation is where the graphs cross. They cross at point (-3, -2).

B) The solutions for f(x) would be points that fall on the graph of f(x). Two possible points are (-3, -2) and (-7, 3)

C) These 2 functions cross at (4, 1). That is the solution.
3 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
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