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Svetach [21]
3 years ago
8

Which of the following points is in the solution set of y

Mathematics
1 answer:
labwork [276]3 years ago
5 0
Where is the picture
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You work 40 hours per week and spend 5 hours per week commuting to and from work. Your friend works 56 hours per week and spends
Burka [1]

Answer:

the answer is yes

Step-by-step explanation:

your ratio..40:5

friends ratio..56:7

each quantity is 8

3 0
3 years ago
Which liner function represents the line given by the points-slope equation y-8=1/2(x-4)?
Gelneren [198K]
<span>y = (1/2)x + 6. hope it helps
</span>
3 0
3 years ago
Which relation is a function? Help please
WARRIOR [948]

Answer:

The bottom right (v shaped one) is a function

Step-by-step explanation:

Hope this helps:)

4 0
2 years ago
Read 2 more answers
Use the law of cosines to find the value of 2*4*5 cos theta
sweet-ann [11.9K]
We can proceed in solving the problem since all information are given such as 2*4*5costheta.
we have a=4, b=5, and C=theta
let us solve for "c" using Pythagorean
c²=a²+b²
c²=4²+5²
c=6.4
Solving for theta or C
c²=a²+b²-2abcosC
6.4²=4²+5²-2*4*5*cosC
C=90

3 0
3 years ago
Read 2 more answers
What is the sum of the first 51 consecutive odd positive integers?
Angelina_Jolie [31]
We call:

a_{n} as the set of <span>the first 51 consecutive odd positive integers, so:

</span>a_{n} = \{1, 3, 5, 7, 9...\}

Where:
a_{1} = 1
a_{2} = 3
a_{3} = 5
a_{4} = 7
a_{5} = 9
<span>and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

</span>a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}
<span>
Then:

</span>a_{n} = a_{1} + (n-1)d
<span>
So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

</span>S_{k} = ( \frac{a_{1} +  a_{k}}{2}  ).k
<span>
Therefore, we need to find:
</span>a_{k} =  a_{51}  

Given that a_{1} = 1, then:

a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1

Thus:
a_{k} = a_{51} = 2(51)-1 = 101

Lastly:

S_{51} = ( \frac{1 + 101}{2} ).51 = 2601 

4 0
3 years ago
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