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2 years ago

The set of ordered pairs below represents a function.

Mathematics

1 answer:

ArbitrLikvidat [17]2 years ago

8 0

Given:

The set of ordered pairs below represents a function.

{(–9, 1), (–5, 2), (0, 7), (1, 3), (6, –10)}

Josiah claims that the ordered pair (6, –10) can be replaced with any ordered pair and the set will still represent a function.

To find:

All the ordered pairs that could be used to show that this claim is incorrect.

Solution:

We need to find the ordered pairs that can be replaced with (6,-10) and for which the set is not a function.

A relation is called function if there exist a unique output for each input.

If (-9,-6) is replaces with (6,-10), then the set has two outputs y=1 and y=-6 for x=-9. So, the claim is incorrect.

If (1,12) is replaces with (6,-10), then the set has two outputs y=12 and y=3 for x=1. So, the claim is incorrect.

If (-10,-10) is replaces with (6,-10), then there exist unique output for each input. So, the claim is correct.

If (4,7) is replaces with (6,-10), then there exist unique output for each input. So, the claim is correct.

If (-5,1) is replaces with (6,-10), then the set has two outputs y=1 and y=2 for x=-5. So, the claim is incorrect.

Therefore, the correct options are (a), (b) and (e).

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Write the equation of the line that passes through (-6,0) and is parallel to the line y=5x-43

AVprozaik [17]

The equation of the line that passes through (-6,0) and is parallel to the line y = 5x - 43 in slope intercept is y = 5x + 30

Solution:

Given that line that passes through (-6, 0) and is parallel to the line y = 5x - 43

We have to find the equation of line

Let us first find slope of line having equation y = 5x - 43

The slope intercept form of line is given as:

y = mx + c ------- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing equation of line y = 5x - 43 with slope intercept form y = mx + c,

we get m = 5

Thus slope of given line is 5

We know that slopes of parallel lines are equal

Thus the slope of line parallel to given line is also 5

Now let us find equation of line having slope "5" and passes through (-6, 0)

Substitute m = 5 and (x, y) = (-6, 0) in eqn 1

y = mx + c

0 = 5(-6) + c

Thus the required equation is:

Substitute c = 30 and m = 5 in eqn 1

y = 5x + 30

Thus the required equation of line is found

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3 years ago

Find 2004-04-02-04-00_files/i0340000.jpg. Round the answer to the nearest tenth.

lesya692 [45]

The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them

we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
cos C=[a²+b²-c²]/[2*a*b]---> cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°

applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°

calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°

the angles of the triangle are
A=54.6°
B=79.3°
C=46.1°

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3 years ago

Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?

algol [13]

We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points. That will be the slope of the tangent to the curve. Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.

It's not written as a function, but it can still be differentiated using the chain rule:

x2 + xy + y2 = 3

(2x) + (x dy/dx + y dx/dx) + (2y dy/dx) = 0

(I used parentheses to show the differentiation of each term in the original equation.)

2x + x dy/dx + y + 2y dy/dx = 0

2x + y = -x dy/dx - 2y dy/dx

2x + y = dy/dx (-x -2y)

-(2x + y)/(x + 2y) = dy/dx

We have the derivative of y, but it's defined partly in terms of y itself. That's OK. Let's go on...

So where would the slope be infinite? That would happen when x + 2y = 0, or y = -x/2

Let's plug that in for y in the original equation to find points where that's the case.

x2 + xy + y2 = 3

x2 + x(-x/2) + (-x/2)2 = 3

x2 - x2/2 + x2/4 = 3

3x2 / 4 = 3

x2 = 4

x = ±2

So we have two x values where the tangent might be vertical. Let's plug them into the equation and see what the y values are. First x = 2...

x2 + xy + y2 = 3

4 + 2y + y2 = 3

y2 + 2y + 1 = 0

(y + 1)2 = 0

y = -1

So at the point (2, -1) the tangent is vertical.

Now try x = -2...

x2 + xy + y2 = 3

4 - 2y + y2 = 3

y2 - 2y + 1 =0

(y - 1)2 = 0

y = 1

So at the point (-2, 1) the tangent is vertical.

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2 years ago

The table shows how an elevator 500 feet above the ground is descending at a steady rate.

lapo4ka [179]

Answer:

h(t) = -5t + 500

Step-by-step explanation:

At time zero, the elevator is at 500 ft.

At time 5 seconds, the elevator is at 475 ft.

The difference in height is -25 ft.

The elevator traveled -25 ft (25 ft down) in 5 seconds, so it travels at a speed of 5 ft per second. Since it is traveling down, its speed is -5 ft/sec.

For each second of travel, it loses 5 ft of height.

h(t) = -5t + 500

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2 years ago

PLEASE HELP!!!!!!!! WILL MARK AS BRAINLIEST!!!!!

Ratling [72]

A.) 2, 7

and heres other words so i can send it

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2 years ago