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2 years ago

Counter example. The sum of three fractions with odd numerators is never 1/2

Mathematics

2 answers:

kolbaska11 [484]2 years ago

8 0

1/6 + 1/6 + 1/6 = 3/6 = 1/2

tia_tia [17]2 years ago

5 0

One <em>possible counterexample</em> is:

1/4 + 1/5 + 1/20 = 1/2.

In this problem, we would find the least common denominator for 4, 5 and 20. The first thing all 3 numbers will divide into is 20:

1/4 = 5/20; 1/5 = 4/20; 1/20 = 1/20

This gives us: 5/20 + 4/20 + 1/20

Adding the numerators, we get (5+4+1)/20 = 10/20 = 1/2.

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Jordan used the distributive property to write an expression that is equivalent to 6c – 48. 6c - 48 is equivalent to 6(c - 48) I

Katyanochek1 [597]

It’s incorrect.

The correct answer is 6(c - 8)

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2 years ago

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What number comes next 16, 06, 68, 88,

inessss [21]

So the given series is "16, 06, 68, 88, __"

Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).

From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.

Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.

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2 years ago

How many marbles, each with a volume of 36 cubic centimeters, are needed to fill in a cylindrical vase with a radius of 6 centim

ANEK [815]

The volume of a cylinder can be found using the formula:

π r² h,

where r is the radius of the circular base and h is the height of the cylinder.

If we plug in the measurements of the cylinder, we get:

π (6²) (28)

When this is simplified, we get that the volume of the cylinder is:

1008π cubic cm

Thus, if each marble has a volume of 36π cubic cm, then to find how many marbles will fit into the vase we must divide the vases total volume by the volume of each marble.

1008π / 36π = 28

Therefore, the answer is D. 28 marbles

7 0

2 years ago

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Ronald spins a spinner that has 10 sections numbered from 1 to 10. What is the probability that the spinner lands on a prime num

kobusy [5.1K]

The probability the spinner lands on a prime number is 4/10 or 40%

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2 years ago

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WILL GIVE BRAINLIEST AND 39 POINTS

Mashcka [7]

Part 1)
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hours, the velocity of the car is 59 km/h.

Part 1 a): Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used

Let

A(3,51) B(5,59)

x------ > represent different times

y------ > represent the velocity of the car

Step 1

Find the slope AB

m=(y2-y1)/(x2-x1)------ > m=(59-51)/(5-3)------ > m=8/2---- > m=4

Step 2

With m=4 and point A(3,51) find the equation of the line

y-y1=m*-(x-x1)------ > y-51=4*(x-3)----- > y=4x-12+51----- > y=4x+39

we know that

The standard form of line equation is Ax + By = C

y=4x+39----- > y-4x=39------ > this is the standard form

the answer part 1 a) is

y-4x=39

Part 1 b) How can you graph the equation obtained in Part a) for the first six hours?

To graph the equation obtained in Part a) plot the point A and the point B
and join the points to draw the line

To obtain the velocity for the first six hours, substitute the value of x=6 hour in the equation

for x=6 hour

y-4x=39------ > y-4*6=39------ > y=39+24------ > y=63 km/h

using a graph tool

see the attached figure N 1

Part 2)

g(x)=1+1.5^x

step 1

find the equation of the line of f(x)

let

A(-5,3) B(-3,-1)

m=(-1-3)/(-3+5)----- > m=-4/2---- > m=-2

with m=-2 and point A

y-y1=m*(x-x1)------ > y-3=-2*(x+5)---- > y=-2x-10+3----- > y=-2x-7

f(x)=-2x-7

step 2

find the equation of the line of p(x)

let

C(0,2) D(-2,-3)

m=(-3-2)/(-2-0)----- > m=-5/-2---- > m=2.5

with m=2.5 and point C

y-y1=m*(x-x1)------ > y-2=2.5*(x-0)---- > y=2.5x+2

p(x)=2.5x+2

Part 2 a) What is the solution to the pair of equations represented by p(x) and f(x)?

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 2

The solution is the point (-2,-3)

Part 2 b) Write any two solutions for f(x).

f(x)=-2x-7

for x=0

f(0)=2*0-7---- > f(0)=-7

solution 1 is the point (0,-7)

for x=1

f(1)=2*1-7---- > f(1)=-5

solution 2 is the point (1,-5)

Part 2 c) What is the solution to the equation p(x) = g(x)?

We have

p(x)=2.5x+2

g(x)=1+1.5^x

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 3

The solution are the points (0,2) and (7.3,20.2)

Part 3
)

Part A:There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.

An example of such system of equation is

x < 0

y > 0

The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.

see the attached figure N 4

Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.

Substituting D(-4, 2) into the system

we have:

-4 < 0

2 > 0

as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.

Also,

substituting E(-1, 5) into the system we have:

-1 < 0

5 > 0

as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.

To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.

For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true

For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false

For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false

For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true

For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true

For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true

Therefore,

the farms in which chicken can be raised are the farms at point A, D, E and F.

5 0

2 years ago

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