3 years ago

A person walks 1/7

Mathematics

1 answer:

balandron [24]3 years ago

6 0

Answer:

3 Miles per hour. (MPH)

Step-by-step explanation:

This would be 1/7 x 21 = 21/7 = 3

You might be interested in

7/8 4/8 how much farther does 7/8 have than 4/8

MaRussiya [10]

The Correct Answer Is 3/8

6 0

3 years ago

What is the 9th term in the sequence 2, -6, 18, -54, ...?

gregori [183]

Your answer would be 13,122

You would multiply by -3 each time for this particular sequence.

Hope I helped!

6 0

3 years ago

The admission fee at an amusement park is $1.75 for children and $4.80 for adults. On a certain day, 303 people entered the park

Nookie1986 [14]

Answer:

Children 188

Adults. 115

Step-by-step explanation:

Let the no. of children be x and adults be y

x + y = 303

x = 303 - y. .... .....(1)

1.75x + 4.80y = 881. ...........(2)

Substituting,

1.75(303-y) +4.80y = 881

530.25 -1.75y + 4.80y = 881

530.25 + 3.05y = 881

3.05y = 881 - 530.25

y = 350.75 / 3.05 = 115 = adults

Children = 303-115 = 188

3 0

3 years ago

Type the correct answer in each box. If necessary, use / for the fraction bar. A bag contains 5 blue marbles, 2 black marbles, a

serg [7]

Answer:

P (not drawing a black marble) = 4/5

P (red marble) = 3/10

Step-by-step explanation:

The total number of marbles is

5 blue marbles +2 black marbles+ 3 red marbles = 10 marbles

P (not drawing a black marble) =

The marbles that are not black/ total marbles

(blue + red)/ total

(5+3)/10

8/10

4/5

P (red marble) = red marble/ total

= 3/10

7 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago