Which of the relations given by the following sets of ordered pairs is a function?
o {(5,2), ( - 4, 2), (3,6), (0,4), (- 1, 2)}
o {(5, 4), (5, 6), (5,8), (5, 10), (5, 12)}
{(-3, - 2), ( - 2, – 1), (0, - 1), (0, 1), (1, 2)}
{(7,3), ( – 6,8), ( – 3,5), (0, – 3), (7, 11)}

Answers

Answer 1

9514 1404 393

Answer:

  (a)  {(5, 2), (-4, 2), (3, 6), (0, 4), (-1, 2)}

Step-by-step explanation:

The only relation with no repeated x-values is the first one. The first relation is a function.


Related Questions

[18].Simplify (TTE): x(2x+y+5) - 2(x²+xy+5) + y(x + y)

Answers

Answer:

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 5x -10 + y\²[/tex]

Step-by-step explanation:

Given

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y)[/tex]

Required

Simplify

We have:

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y)[/tex]

Open brackets

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 2x\²+xy+5x - 2x\²-2xy-10 + xy + y\²[/tex]

Collect like terms

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 2x\²- 2x\²+xy-2xy+ xy+5x -10 + y\²[/tex]

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 5x -10 + y\²[/tex]

Three numbers form an arithmetic sequence whose
common difference is 3. If the first number is
increased by 1, the second increased by 6, and
the third increased by 19, the resulting three
numbers form a geometric sequence. Determine
the original three numbers.

Answers

Let x be the first number in the sequence. Then the first three numbers are

{x, x + 3, x + 6}

The next sentence says that the sequence

{x + 1, x + 9, x + 25}

is geometric, which means there is some fixed number r for which

x + 9 = r (x + 1)

x + 25 = r (x + 9)

Solve for r :

r = (x + 9)/(x + 1) = (x + 25)/(x + 9)

Solve for x :

(x + 9)² = (x + 25) (x + 1)

x ² + 18x + 81 = x ² + 26x + 25

8x = 56

x = 7

Then the three numbers are

{7, 10, 13}

Find the area of a triangle as a mixed number.

Answers

Answer:

I believe the answer is 4 37/50!

In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that a student who has a cat also has a dog?
Has a cat Does not have a cat
Has a dog 7 6
Does not have a dog 8 2

Answers

Outcome C joint D = 7, D excluding C is 6, C excluding D = 8, no C no D =2. Should be 23 people in total in that class. 7/23.

35 + 3 x n with n = 7

Answers

56 would be the answer
It would just be 56 you can plug this in a Calcator

e movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. How could you correctly rewrite the equation 4(10+5) = 6(12 - 2) using the distributive property​

Answers

9514 1404 393

Answer:

  4·10 +4·5 = 6·12 -6·2

Step-by-step explanation:

Each outside factor multiplies each inside term.

  4(10 +5) = 6(12 -2)

  4·10 +4·5 = 6·12 -6·2

the guys is wrong i checked

Answers

Step-by-step explanation:

the guys is wrong i checked

If we decrease a dimension on a figure, how is the figure’s area affected?
The area decreases.
The area increases.
The area becomes 0.
The area remains the same.

Answers

The area decreases :)

in each figure below find m<1 and m < 2 if a||b

please help i don't have a lot of time I will give brainliest if you help

Answers

Answer:

m∠1 = 105°

m∠2 = 75°

Step-by-step explanation:

From the picture attached,

Two lines 'a' and 'b' are parallel and a transversal 't' is intersecting these lines at two distinct lines.

Therefore, m∠2 = 75° [Corresponding angles measure the same]

m∠1 + m∠2 = 180° [Linear pair of angles are supplementary]

m∠1 + 75° = 180°

m∠1 = 105°

Use the order of operations to simplify 3/4+8(2.50-0.5).

Answers

Answer:

16[tex]\frac{3}{4}[/tex]

Step-by-step explanation:

I need help please can not figure out this problem

Answers

1. (2x-3)(x-4)
2x(x) =2x^2
2x(-4)= -8x
-3(x)= -3x
-3(-4)= 12
2x^2 -8x-3x+12= 2x^2 -11x+12

2. (x+2)(x+5)
x^2+5x+2x+10= x^2 +7x+10

3. (3x-1)(x+2)
3x^2 +6x-x-2= 3x^2+5x-2

A water trough is 9 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 70 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep?

Answers

Answer:

dv = surface area * dh

so

dv/dt = surface area * dh/dt

width at surface = 40 + (80-40)(30/40)

= 40 + 30 = 70 cm = 0.70 m

so

surface area = 9 * .7 = 6.3 m^2

so

.3 m^3/min = 6.3 m^2 * dh/dt

and

dh/dt = .047 meters/min or 4.7 cm/min

Step-by-step explanation:

PLZ ANSWER QUESTION IN PICTURE

Answers

Answer: y = 3x + 6

Step-by-step explanation:

(x-intercept of -2: (-2,0))

(slope = m)

y = mx + b, (-2,0), m = 3

[tex]y=mx+b\\0=3(-2)+b\\0=-6+b\\b=6\\y=3x+6[/tex]

Can someone do #4 and #5

Answers

Answer:

First, find two points on the graph:

(x₁, y₁) = (0, 2)(x₂, y₂) = (2, 8)

Slope = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1}} = \frac{8-2}{2-0} =\frac{6}{2}=3[/tex]

16 + (-3) = 16 - 3 = 13

The smallest positive solution of tan bx = 2 is x = 0.3. Determine the general solution of tan bx = 2.

Answers

The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].

From Trigonometry we remember that Tangent is a Transcendental Function that is positive both in 1st and 3rd Quadrants and have a periodicity of [tex]\pi[/tex] radians.  The procedure consists in using concepts of Direct and Inverse Trigonometric Functions as well as characteristics related to the behavior of the tangent function in order to derive a General Formula for every value of [tex]x[/tex], measured in radians.

First, we solve the following system of equations for [tex]b[/tex]:

[tex]\tan bx = 2[/tex] (1)

[tex]x = 0.3[/tex] (2)

Please notice that angles are measured in radians.

(2) in (1):

[tex]\tan 0.3b = 2[/tex]

[tex]0.3\cdot b = \tan^{-1} 2[/tex]

[tex]b = \frac{10}{3}\cdot \tan^{-1}2[/tex]

[tex]b\approx 3.690[/tex]

Under the assumption of periodicity, we know that:

[tex]y = \tan bx[/tex]

[tex]b\cdot x \pm \pi\cdot i = \tan^{-1} y[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

[tex]b\cdot x = \tan^{-1}y \mp \pi\cdot i[/tex]

[tex]x = \frac{\tan^{-1}y \mp \pi\cdot i}{b}[/tex]

If we know that [tex]y = 2[/tex] and [tex]b \approx 3.690[/tex], then the general solution of this trigonometric function is:

[tex]x = \frac{0.352\pi \mp \pi\cdot i}{3.690}[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

[tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].

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Find the derivative on the value of x=-4
[tex]y=(6x-5)\sqrt{8x-3}[/tex]

Answers

[tex]\\ \sf\longmapsto y=(6x-5)\sqrt{8x-3}[/tex]

[tex]\\ \sf\longmapsto y=6(-4)-5\sqrt{8(-4)-3}[/tex]

[tex]\\ \sf\longmapsto y=-24-5\sqrt{-32-3}[/tex]

[tex]\\ \sf\longmapsto y=-29\sqrt{-35}[/tex]

[tex]\\ \sf\longmapsto y=-29\times 35i[/tex]

[tex]\\ \sf\longmapsto y=-1015i[/tex]

strontium-90 is a radioactive material that decays according to the function A(t)=A0e−0.0244t, where A0 is the initial amount present and A is the amount present at time t​ (in years). Assume that a scientist has a sample of 400 grams of​ strontium-90. ​
(a) What is the decay rate of​ strontium-90?
​(b) How much​ strontium-90 is left after 30 ​years?
​(c) When will only 100 grams of​ strontium-90 be​ left?
​(d) What is the​ half-life of​ strontium-90?

​(a) The decay rate of​ strontium-90 is nothing​%.
​(Type an integer or a decimal. Include the negative sign for the decay​ rate.)

Answers

Answer:

Step-by-step explanation:

The decay rate of strontium-90 is -.0244 as given.

For b., we have to use the formula to find out how much is left after 30 years. This will be important for part d.

[tex]A(t)=400e^{-.0244(30)}[/tex] which simplifies a bit to

A(t) = 400(.4809461353) so

A(t) = 192.4 g

For c., we have to find out how long it takes for the initial amount of 400 g to decay to 100:

[tex]100=400e^{-.0244t}[/tex]. Begin by dividing both sides by 400:

[tex].25=e^{-.0244t[/tex] and then take the natural log of both sides:

[tex]ln(.25)=lne^{-.0244t[/tex] . The natural log and the e cancel each other out since they are inverses of one another, leaving us with:

ln(.25) = -.0244t and divide by -.0244:

61.8 years = t

For d., we figured in b that after 30 years, 192.4 g of the element was left, so we can use that to solve for the half-life in a different formula:

[tex]A(t)=A_0(.5)^{\frac{t}{H}[/tex] and we are solving for H. Filling in:

[tex]192.4=400(.5)^{\frac{30}{H}[/tex] and begin by dividing both sides by 400:

[tex].481=(.5)^{\frac{30}{H}[/tex] and take the natural log of both sides, which allows us to pull the exponent out front. I'm going to include that step in with this one:

ln(.481) = [tex]\frac{30}{H}[/tex] ln(.5) and then divide both sides by ln(.5):

[tex]\frac{ln(.481)}{ln(.5)}=\frac{30}{H}[/tex] and cross multiply and isolate the H to get:

[tex]H=\frac{30ln(.5)}{ln(.481)}[/tex] and

H = 28.4 years

give the size of the letter figure below​

Answers

Answer: 150 degrees

Step-by-step explanation: 10+ 20 = 30

180-30 = 150 degrees.

4) The measure of the linear density at a point of a rod varies directly as the third power of the measure of the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center, find the total mass of the given rod and the center of the mass​

Answers

Answer:

a. 16 slug b. 3.2 ft

Step-by-step explanation:

a. Total mass of the rod

Since the linear density at a point of the rod,λ varies directly as the third power of the measure of the distance of the point form the end, x

So, λ ∝ x³

λ = kx³

Since the linear density λ = 2 slug/ft at then center when x = L/2 where L is the length of the rod,

k = λ/x³ = λ/(L/2)³ = 8λ/L³

substituting the values of the variables into the equation, we have

k = 8λ/L³

k = 8 × 2/4³

k = 16/64

k = 1/4

So, λ = kx³ = x³/4

The mass of a small length element of the rod dx is dm = λdx

So, to find the total mass of the rod M = ∫dm = ∫λdx we integrate from x = 0 to x = L = 4 ft

M = ∫₀⁴dm

= ∫₀⁴λdx

= ∫₀⁴(x³/4)dx

= (1/4)∫₀⁴x³dx

= (1/4)[x⁴/4]₀⁴

= (1/16)[4⁴ - 0⁴]

= (256 - 0)/16

= 256/16

= 16 slug

b. The center of mass of the rod

Let x be the distance of the small mass element dm = λdx from the end of the rod. The moment of this mass element about the end of the rod is xdm =  λxdx = (x³/4)xdx = (x⁴/4)dx.

We integrate this through the length of the rod. That is from x = 0 to x = L = 4 ft

The center of mass of the rod x' = ∫₀⁴(x⁴/4)dx/M where M = mass of rod

= (1/4)∫₀⁴x⁴dx/M

= (1/4)[x⁵/5]₀⁴/M

= (1/20)[x⁵]₀⁴/M

= (1/20)[4⁵ - 0⁵]/M

= (1/20)[1024 - 0]/M

= (1/20)[1024]/M

Since M = 16, we have

x' =  (1/20)[1024]/16

x' = 64/20

x' = 3.2 ft

If a tank holds 6000 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as

V=5000 (1-1/50*t)^2 0⤠t ⤠50.

1. Find the rate at which water is draining from the tank after the following amount of time. (Remember that the rate must be negative because the amount of water in the tank is decreasing.)

a. 5 min
b. 10 min
c. 20 min
d. 50 min

2. At what time is the water flowing out the fastest?
3. At what time is the water flowing out the slowest?

Answers

Answer: hello from the question the volume of tank = 6000 gallons while the value in the Torricelli's equation = 5000 hence I resolved your question using the Torricelli's law equation

answer:

1) a) -180  gallons/minute ,

  b)  -160 gallons/minute

  c)  -120 gallons/minute

  d) 0

2) The water is flowing out fastest when t = 5 min

3) The water is flowing out slowest after t = 20 mins

Step-by-step explanation:

Volume of tank = 5000 gallons

Time to drain = 50 minutes

Volume of water remaining after t minutes by Torricelli's law

V = 5000 ( 1 - [tex]\frac{1}{50}t[/tex] )^2 ----- ( 1 )

1) Determine the rate at which water is draining from the tank

First step : differentiate equation 1 using the chain rule to determine the rate at which water is draining from the tank

V' = [tex]-10000[ ( 1 - \frac{1}{50}t ) (\frac{1}{50}) ][/tex]

a) After t = 5minutes

V' = - 10000[ ( 1 - 0.1 ) * ( 0.02 ) ]

   = -180  gallons/minute

b) After t = 10 minutes

V' = - 10000[ ( 1 - 0.2 ) * ( 0.02 ) ]

   = - 160 gallons/minute

c) After t = 20 minutes

V' = - 10000 [ ( 1 - 0.4 ) * ( 0.02 ) ]

   = -120 gallons/minute

d) After t = 50 minutes

V' = - 10000 [ ( 1 - 1 ) * ( 0.02 ) ]

   = 0 gallons/minute

2) The water is flowing out fastest when t = 5 min

3) The water is flowing out slowest after t = 20 mins  because no water flows out after 50 minutes

lim(x-0) (sinx-1/x-1)

Answers

9514 1404 393

Answer:

as written: the limit does not existsin(x-1)/(x-1) has a limit of sin(1) ≈ 0.841 at x=0

Step-by-step explanation:

The expression written is interpreted according to the order of operations as ...

  sin(x) -(1/x) -1

As x approaches 0 from the left, this approaches +∞. As x approaches 0 from the right, this approaches -∞. These values are different, so the limit does not exist.

__

Maybe you intend ...

  sin(x -1)/(x -1)

This can be evaluated directly at x=0 to give sin(-1)/-1 = sin(1). The argument is interpreted to be radians, so sin(1) ≈ 0.84147098...

The limit is about 0.841 at x=0.

write your answer in simplest radical form​

Answers

Answer:

z = √3

Step-by-step explanation:

sin (30°) = z / 2√3

z = sin (30°) 2√3

z = √3

Conan puts tennis balls into tubes after gym class. There are 17 tennis balls, and each tube holds 3 balls. How many tubes does Conan completely fill? How many tennis balls are left?

Answers

the answer is 5 tubes 2 left over

What is the phase of y= -3cos (3x-pi) +5

Answers

Answer:

[tex]- \frac{\pi}{3}[/tex]

Step-by-step explanation:

Given

[tex]y = -3\cos(3x - \pi) + 5[/tex]

Required

The phase

We have:

[tex]y = -3\cos(3x - \pi) + 5[/tex]

Rewrite as:

[tex]y = -3\cos(3(x - \frac{\pi}{3})) + 5[/tex]

A cosine function is represented as:

[tex]y = A\cos(B(x + C)) + D[/tex]

Where:

[tex]C \to[/tex] Phase

By comparison:

[tex]C = - \frac{\pi}{3}[/tex]

Hence, the phase is: [tex]- \frac{\pi}{3}[/tex]

Part b c and d please help

Answers

Answer:

b) Y =5.73X +4.36

C)  =5.73225*(21)X +4.359

    124.73625

D) 163.728 = 5.73X +4.36  

     X = (163.728 - 4.36)/5.73

     X = 27.81291449

  Year would be 2027

Step-by-step explanation:

x1 y1  x2 y2

4 27.288  16 96.075

   

(Y2-Y1) (96.075)-(27.288)=   68.787  ΔY 68.787

(X2-X1) (16)-(4)=    12  ΔX 12

   

slope= 5 41/56    

B= 4 14/39    

   

Y =5.73X +4.36      

(x/4) + (2x/7 =135 solve it​

Answers

Answer:

the ans is 252................

A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 55% salt and Solution B is 70% salt. She wants to obtain 30 ounces of a mixture that is 60% salt. How many ounces of each solution should she use?

Answers

Answer:

Let x = the number of ounces of Solution A

Let y = the number of ounces of Solution B

x + y = 180        y = 180 - x

.60x + .85y = .75(180)

.60x + .85y = 135      Multiply both sides of the equation by 100 to remove the decimal points.

60x + 85y = 13500

60x + 85(180 - x) = 13500

60x + 15300 - 85x = 13500

-25x = -1800

x = 72ounces

y = 180 - 72

y = 108 ounces          

Step-by-step explanation:

Wyzant (ask an expert) solution on their website.

if my savings of $x grows 10 percent each year, how much will i have in 2 years?

Answers

Answer:

$240

Step-by-step explanation:

A year has 12 month in it so lets multiply the 10 by 12 which is $120,Mean a year is $120 so 2years will be $120×2 which is $240

Answer:

x+1/5x

Step-by-step explanation:

Because the eqaution would be x+10%=x+1/10+10%=1/5+x

Then the equation equals x+1/5

From September 1991 to September 1994 the enrollment at a particular school declined by 20 percent. If the number of students enrolled at that school in September 1994 was 720, what was the enrollment in September 1991

Answers

Answer:

900

Step-by-step explanation:

Given that :

Enrollment declined by 20% from between September 1991 to September 1994

This means there was a reduction in enrollment ;

Enrollment in September 1994 = 720

Enrollment in September 1991 = x

Hence,

Enrollment in 1994 = (1 - decline rate ) * enrollment in 1991

720 = (1 - 20%) * x

720 = (1 - 0.2) * x

720 = 0.8x

720 / 0.8 = 0.8x/0.8

900 = x

Hence, Enrollment in September 1991 = 900 enrollments

What is the first step to solve the equation 16x-21 = 52?

1 Add 52 to both sides

2 Add 21 to both sides

3 Subtract 21 from both sides

4 Subtract 52 from both sides

Answers

Answer:

2) Add 21 to both sides

Step-by-step explanation:

When solving [tex]16x-21=52[/tex] for [tex]x[/tex], our goal to isolate [tex]x[/tex] such that we have [tex]x[/tex] set equal to something.

Therefore, we want to start by adding 21 to both sides. This leaves us with [tex]16x=73[/tex] and we are one step closer to isolating [tex]x[/tex].

subtract 52 from both sides
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