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2 years ago

Solve the equation (5×+7)1/5=(8-6×)1/5

1 answer:

5 0

(5x+7)(1/5)=(8−6x)(1/5)
Simplify both sides. In order to do this you need to distribute 1/5 in each expression.
(5x)(1/5)+(7)(1/5)=(8)(1/5)+(−6x)(1/5)
x+7/5=8/5+−6/5x
x+7/5=−6/5x+8/5
Now you need to add (-6/5)x to both sides. This leaves you with
(5/11)x+7/5=8/5
Now subtract 7/5 from both sides.
(5/11)x=1/5
Divide both sides by 5/11
x=1/11

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H = {(2, 5), (3, 5), (4, 5), (5, 5), (6, 5)} Is H a function and why?.

mixer [17]

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Yes, each x-value has a unique y-value.

Step-by-step explanation:

This is a function because each x-value has its own y-value. If this is not the case then it is not a function, because then, two points with the same x-value and different y-values would fail the VLT.

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The vertical line test (VLT) is a simple method that mathematicians made because they were lazy to make a table of values and find duplicates.

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1 year ago

30 men repair a road in 56 days by working 6 hours daily. in how many days 45 men will repair same road by working 7 hours daily

zepelin [54]

Answer:

Case1:

Men : 30

Work : 1

Time (day×hr): 56×6 = 336 hr.

Case2:

Men : let it be m men.

Work: 1

Time: 45×7 = 315 hr.

Work being constant in both cases, men and time are in inverse proportion i.e, more men take less time.

Product of men and time is constant in both cases.

Therefore, 30×336=m×315

Or, 30×336/315 = m

Or, m = 32.

Hence, required number of men is 32.

Step-by-step explanation:

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2 years ago

There are 100 students at a college who have taken a course in calculus, 110 who have taken a course in discrete mathematics, an

vesna_86 [32]

Answer:

110 students

Step-by-step explanation:

The number of students who have only taken calculus is given by the number of students who have taken calculus minus the students who have taken both classes:

$C =100 -50\\C= 50\ students$

The number of students who have only taken discrete mathematics given by the number of students who have taken discrete mathematics minus the students who have taken both classes:

$D =110 -50\\C= 60\ students$

The number of students that have taken a course in either calculus or discrete mathematics is:

$E = C+D = 50+60\\E= 110\ students$

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