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Veronika [31]
3 years ago
8

30 points And i will give brainliest

Mathematics
2 answers:
julia-pushkina [17]3 years ago
4 0

Answer:

The one with the frequency of 8

Step-by-step explanation:

viktelen [127]3 years ago
3 0
Hey is right do as he says
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Hi :)<br> Can anyone help me out?
elena55 [62]

Answer:

number 5:

(0 minus 1) minus 3= negative 4

number 6:

(0-4)+3=-1

5 0
3 years ago
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What is the equation of the line
Inga [223]
Y= 1/2x + 2 is the correct answer
4 0
3 years ago
Eduardo thinks of a number between 1 and 20 that has exactly 5 factors .what number is he thinking about
Natali5045456 [20]

Answer

16

Explanation

Factors are number that you can multiply to get another number. A number between 1 and 20 that has many factors are 12, 16, 18..

We can find the factors of this numbers and identify the one with five factors.

12→1,2,3,4,6,and 12

16→1,2,4,8,and 16

18→1,2,3,6,9,and 18

We find the 16 has exactly 5 factors.  

Read more on Brainly.com - brainly.com/question/195242#readmore

4 0
3 years ago
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The coefficient in the expression -x - 10 is
aivan3 [116]

Answer: I think the coefficient is -x

8 0
3 years ago
Read 2 more answers
Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisti
timofeeve [1]

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

\bf \int_{C}F

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write  

C = \bf C_1+C_2

where  

\bf C_1 =  line segment from (1, 5) to (0, 0)  

\bf C_2 = line segment from (0, 0) to (0, 9)

so,

\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

<em>r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1 </em>

Hence \bf C_1 can be parameterized as

\bf r_1(t) = (1-t, 5-5t) for 0 ≤ t ≤ 1

and \bf C_2 can be parameterized as

\bf r_2(t) = (0, 9t) for 0 ≤ t ≤ 1

The derivatives are

\bf r_1'(t) = (-1, -5)

\bf r_2'(t) = (0, 9)

and

\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0

\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0

In consequence,

\bf \int_{C}F=0

6 0
4 years ago
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