Which of the following graphs represent a function? 4072-01-02-01- a. Graph A and Graph C b. Graph A c. Graph D d. Graph B and Graph D
Answers
The graph which represent a graph of a function is:
Graph A and Graph C
Step-by-step explanation:We know that a graph of a function satisfies the vertical line test i.e. any line passing through the domain and parallel to y-axis should intersect the curve exactly once i.e. corresponding to each x-value there is exactly one y-value.
Hence, from the figure attached to the answer we see that the Graph which is a function is:
Graph A and Graph C
Graph A represents a function.
Explanation:A function is a relation in which each input has only one output. In other words, for every x-value, there can only be one y-value. To determine if a graph represents a function, we can use the vertical line test. If a vertical line intersects the graph at more than one point, then the graph does not represent a function. Looking at the given graphs, Graph B and Graph D do not pass the vertical line test, as there are vertical lines that intersect these graphs at multiple points. Therefore, the correct answer is Graph A.
Related Questions
D is the midpoint of CE.E has coordinates (-3,-2), and D has coordinates (2 1/2, 1). Find the coordinate of C.
Answers
(-3,-2)....x1 = -3 and y1 = -2
(x,y)......x2 = x and y2 = y
now we sub
(-3 + x) / 2, (-2 + y) / 2
(-3 + x) / 2 = 2 1/2
-3 + x = 5/2 * 2
-3 + x = 5
x = 5 + 3
x = 8
(-2 + y) / 2 = 1
-2 + y = 1 * 2
-2 + y = 2
y = 2 + 2
y = 4
so the coordinates of point C are : (8,4)
The coordinate of point C on line CE will be (8, 4).
What is Coordinates?
A pair of numbers which describe the exact position of a point on a cartesian plane by using the horizontal and vertical lines is called the coordinates.
Given that;
D is the midpoint of CE.
E has coordinates (-3,-2), and D has coordinates (2 1/2, 1).
Now, By the definition of midpoint;
Let the coordinate of point C = (x, y)
Then,
((x + (-3))/2 , (y + (-2))/2) = (2 1/2, 1)
By comparison we get;
x + (-3) / 2= 2 1/2
x - 3 = 2 (5/2)
x - 3 = 5
x = 3 + 5
x = 8
And, (y + (-2))/2 = 1
y - 2 = 2
y = 2 + 2
y = 4
Thus, The coordinate of point C = (x, y) = (8, 4)
So, The coordinate of point C on line CE will be (8, 4).
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Given sina=6/7 and cosb=-1/6, where a is in quadrant ii and b is in quadrant iii , find sin(a+b) , cos(a-b) and tan(a+b)
Answers
now, keep in mind that, the hypotenuse is just a radius unit, and thus is never negative, so if a fraction with it is negative, is the other unit. A good example of that is the second fraction here, -1/6, where the hypotenuse is 6, therefore the adjacent side is -1. Anyhow, let's find the opposite side to get the sin(b).
[tex]\bf cos(b)=\cfrac{\stackrel{adjacent}{-1}}{\stackrel{hypotenuse}{6}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a \\\\\\ \textit{now, angle "b" is in the III quadrant, where the opposite is negative} \\\\\\ -\sqrt{35}=b\qquad \qquad \boxed{sin(b)=\cfrac{-\sqrt{35}}{6}}[/tex]
now
[tex]\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ [/tex]
[tex]\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\qquad tan({{ \alpha}} - {{ \beta}}) = \cfrac{tan({{ \alpha}})- tan({{ \beta}})}{1+ tan({{ \alpha}})tan({{ \beta}})}[/tex]
[tex]\bf sin(a+b)=\cfrac{6}{7}\cdot \cfrac{-1}{6}+\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{-1}{7}+\cfrac{\sqrt{455}}{42} \\\\\\ \cfrac{-6+\sqrt{455}}{42}\\\\ -------------------------------\\\\ cos(a-b)=\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-1}{6}+\cfrac{6}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{\sqrt{13}}{42}-\cfrac{\sqrt{35}}{7} \\\\\\ \cfrac{\sqrt{13}-6\sqrt{35}}{42}[/tex]
[tex]\bf -------------------------------\\\\ tan(a)=\cfrac{\frac{6}{7}}{-\frac{\sqrt{13}}{7}}\implies -\cfrac{6}{\sqrt{13}}\implies -\cfrac{6\sqrt{13}}{13} \\\\\\ tan(b)=\cfrac{\frac{-\sqrt{35}}{6}}{\frac{-1}{6}}\implies -\sqrt{35}\\\\ -------------------------------\\\\[/tex]
[tex]\bf tan(a+b)=\cfrac{-\frac{6}{\sqrt{13}}-\sqrt{35}}{1-\left( -\frac{6}{\sqrt{13}} \right)\left( -\sqrt{35} \right)}\implies \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{1-\frac{6\sqrt{35}}{\sqrt{13}}} \\\\\\ \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{\frac{\sqrt{13}-6\sqrt{35}}{\sqrt{13}}}\implies \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}[/tex]
and now, let's rationalize the denominator of that one, hmmm let's see
[tex]\bf \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}\cdot \cfrac{\sqrt{13}+6\sqrt{35}}{\sqrt{13}+6\sqrt{35}} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-\sqrt{5915}-6\sqrt{15925}}{({\sqrt{13}-6\sqrt{35}})({\sqrt{13}+6\sqrt{35}})} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-13\sqrt{35}-210\sqrt{13}}{(\sqrt{13})^2-(6\sqrt{35})^2} \\\\\\ \cfrac{-216\sqrt{13}-49\sqrt{35}}{13-210}\implies \cfrac{-216\sqrt{13}-49\sqrt{35}}{-197} \\\\\\ \cfrac{216\sqrt{13}+49\sqrt{35}}{197}[/tex]
sin(a+b) = -1/7 +√455/42 = 0.8721804464845457
cos(a-b) = √13/42 - √35/7 = -0.7761476987942811
tan(a+b )= (6√13/13 + √35) / (1 - 6√455/13) = -0.525
Given sin(a) = 6/7 and cos(b) = -1/6, with a in quadrant II and b in quadrant III, we need to utilize trigonometric identities to find sin(a+b), cos(a-b), and tan(a+b).
Firstly, since a is in quadrant II, cos(a) is negative. We use the identity sin²(a) + cos²(a)=1 to find cos(a):
cos(a) = -√(1 - sin²(a)) = -√(1 - (6/7)²) = -√(1 - 36/49) = -√(13/49) = -√13/7Similarly, since b is in quadrant III, sin(b) is also negative. We use the identity sin²(b) + cos²(b)=1 to find sin(b):
sin(b) = -√(1 - cos²(b)) = -√(1 - (-1/6)²) = -√(1 - 1/36) = -√(35/36) = -√35/6Now we can use the angle addition and subtraction formulas:
1. sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin(a + b) = (6/7)(-1/6) + (-√13/7)(-√35/6) = -1/7 + √(13×35)/(7×6) = -1/7 + √455/42 = -1/7 +√455/422. cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
cos(a - b) = (-√13/7)(-1/6) + (6/7)(-√35/6) = √13/(7×6) - (6√35)/(7×6) = √13/42 - √35/73. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
Using tan(a) = -sin(a)/cos(a) = -(6/7)/(-√13/7) = 6/√13 and tan(b) = sin(b)/cos(b) = (-√35/6)/(-1/6) = √35:tan(a + b) = (6/√13 + √35) / (1 - (6/√13)(√35)) = (6√13/13 + √35) / (1 - 6√455/13)$185 DVD player 6% markup
Answers
i think.....
if it is right dont forget to give me brainliest
"unable to determine because x can't take on the value 5.5"
Answers
f(x) = -------------------
x - (5.5)
is an example where there's no finite limit if x approaches 5.5.
The diagram represents a reduction of a triangle by using a scale factor of 0.8.
What is the height of the reduced triangle?
4.0 inches
4.8 inches
5.2 inches
7.5 inches
Answers
The reduction factor of 0.8 means the lengths in the reduced triangle are 0.8 times those of the original.
Then the original 6 inch length is reduced to 0.8×6 inches = 4.8 inches in the reduced triangle.
Answer:
4.8 inches
Step-by-step explanation:
The scale factors are used to convert a figure into another one with similar characteristics but different lengths, in this example is a triangle, and in order to calculate the measure of the height you just have to multiply the original height by the scale factor:
6 inches * scale factor
6 inches* 0.8= 4.8 inches
So the resultant triangle will have a height of 4.8 inches.
what 5x(4+gy) if x= 1 g= 20 Y=14
please show work
Answers
[tex]5x(4+gy)[/tex]
[tex]5(1)(4+(20)(14))=5(4+280)=5(284)=1420[/tex]
What's 24.67 to one significant figure?
Answers
Can I have help on number 16 and please explain how you got the answer
Answers
[tex]3c+3(4)\\ 3c+12[/tex]
So the area of this particular rectangle would be 3c + 12 sq. ft
4r+3=r+21
How do I do this?
Answers
4r+3-3=r+21-3
4r=r+18
4r-r=r-r+18
3r=18
3r/3=18/3
r=6
remember that what you do in one side, do it in the other
Reorder the terms 3 + 4r = 21 + r. Solve: 3 + 4r = 21 + r.
Move all terms containing r to the left, all other terms to the right.
Solving for variable 'r'.
Add '-1r' to each side of the equation. 3 + 4r + -1r = 21 + r + -1r.
Combine like terms: 4r + -1r = 3r3 + 3r = 21 + r + -1r
Combine like terms: r + -1r = 0 3 + 3r = 21 + 0 3 + 3r = 21
Add '-3' to each side of the equation. 3 + -3 + 3r = 21 + -3
Combine like terms: 3 + -3 = 0 0 + 3r = 21 + -3 3r = 21 + -3
Combine like terms: 21 + -3 = 18 3r = 18
Divide each side by '3'. r = 6 Simplifying r = 6
605 mi in 11 hours at the same rate how many miles would he drive in 13 hours
Answers
605/ 11= 55
55*13=715
In 13 hours, 715 miles will be traveled
Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(t) = at2 + bt + c, where a, b, and c are to be determined.
Answers
[tex]x=at^2+bt+c[/tex]
[tex]x'=2at+b[/tex]
[tex](2at+b)+2(at^2+bt+c)=t^2+4t+7[/tex]
[tex]2at^2+(2a+2b)t+(b+2c)=t^2+4t+7[/tex]
[tex]\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4[/tex]
Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:
[tex]x'+2x=0\implies r+2=0\implies r=-2[/tex]
so that [tex]x_c=Ce^{-2t}[/tex] is a characteristic solution to the ODE, and the general solution would be
[tex]x=Ce^{-2t}+\dfrac{t^2}2+\dfrac{3t}2+\dfrac{11}4[/tex]
To find a, b, c for the solution:
Let's start by writing down the expression for the function x(t) and its derivative:
We have:
x(t) = at² + bt + c
and
x'(t) = 2at + b
Using x' and x into the differential equation x′ + 2x = t² + 4t + 7 gives us:
2at + b + 2*(at² + bt + c) = t² + 4t + 7
Expanding this gives:
2at² + 2bt + b + 4at + 2c = t² + 4t + 7
By equating the coefficients of equivalent powers of t on both sides, we get three equations:
For t² :
2a = 1
So, a = 1/2
For t:
2b + 4a = 4
Substitute a = 1/2 into the equation gives b = 1 - 2 = -1
For the constant term:
b + 2c = 7
Substituting b = -1 gives c = 4.
So the solution is a = 1/2, b = -1, c = 4.
So the specific solution of this differential equation is given by x(t) = (1/2)t² - t + 4.
Round 5836197 to the nearest hundred
Answers
Answer:
5836200.
Step-by-step explanation:
Given : 5836197 .
To find : Round 5836197 to the nearest hundred.
Solution : We have given 5836197
Step 1 : First, we look for the rounding place which is the hundreds place.
Step 2 : Rounding place is 97.
Step 3 : 97 is greater than 50 then it would be rounded up mean next number to 97 would be increase to 1 and 97 become 00.
Step 4 : 5836200.
Therefore, 5836200.
A, B, and C are polynomials, where A = n, B = 2n + 6, and C = n2 – 1. What is AB – C in simplest form? A=–n2 + 3n + 5 B=n2 + 6n + 1 C=2n2 + 6n – 1 D=3n2 + 5
Answers
Multiplying out the first term: 2n^2 + 6n
Subtracting n^2-1: - (n^2 -1)
----------------------
n^2 + 6n + 1 (answer)
Answer:
B
Step-by-step explanation:
If the radius of a circle measures 2 inches, what is the measure of its diameter?
Answers
~~Alexis
what is the answer of 9X23+3X39-28=n
Answers
n= 296
Answer: n=296
Step-by-step explanation: 9 x 23 + 3 x 39 - 28 + n = 296
Hope this helps! <3
write a fraction less than 1 with a denominator of 6 that is greater than 3/4
Answers
Hope this helps!
Answer:= 5/6
Step-by-step explanation:hope this helps
How do you figure out what 1/10 of 1,7000.000 km squared is?
Answers
This would get you 28,900,000.
how can I adjust a quotient to solve a division problem
Answers
What is the expanded notation for 2.918?
Answers
I hope this helps.
Rochelle found the quotient of an integer, x, and 13. Is her quotient a rational number
Answers
The integers (abbreviated Z) contain all of the natural numbers, and add negative numbers, too (-1, -2, -3, etc.). Multiplication, addition, and subtraction work fine now, and no matter which integers we use, those operations will always give us back another integer. It's not all perfect, though; division gives us a bit of a hassle now. Expressions like 6/3 are fine, since they just produce another integer (2, in this case), but there's no integer result for something like 2/5 or 3/8, so again, we have to create a new set of numbers to compensate.
That set of numbers is what we refer to as the rational numbers (abbreviated Q), named after the fact that all of them are expressed as ratios. By expanding our number system with these, we gain the ability to express every integer as a rational number by putting it over 1 (6 = 6/1, 5 = 5/1, etc.).
So, getting back to the original question, "is the quotient of an integer x and 13 a rational number?" Yes. The quotient of two integers can always be represented as a rational number.
What can be used as a reason in a two-column proof?
Select each correct answer.
conjecture
postulate
definition
premise
Answers
Therefore, what can be used as a reason in a two-column proof are:
Postulates
Definitions
Answer:
Two column proofs are organized into statement and reason columns. Each statement must be justified in the reason column. The reason column will typically include "given", vocabulary definitions, and theorems.
Therefore, what can be used as a reason in a two-column proof are:
Postulates
Definitions
Simplify each expression (combine like terms)
17x+12-13x
-35+17-3w+12w
Answers
And for the second one - 18+9w
Suppose you have two credit cards. The first has a balance of $415 and a credit limit of $1,000. The second has a balance of $215 and a credit limit of $750. What is your overall credit utilization?
Answers
Compute for the total balance:
total balance = $415 + $215 = $630
Then we compute for the total credit limit:
total credit limit = $1,000 + $750 = $1,750
The credit utilization would simply be the percentage ratio of total balance over total credit limit. That is:
credit utilization = ($630 / $1,750) * 100%
credit utilization = 36%
Every two years, a manufacturer's total yearly sales decline 20%. She sells 400 total units in her first year. Rounded to the nearest unit, how many total units will she sell her ninth year?
Answers
we know, the first year she sold 400 units, thus year 0, 400 units, t = 0, A = 400.
[tex]\bf \textit{Periodic Exponential Decay}\\\\ A=I(1 - r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\to &400\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &0\\ p=period \end{cases} \\\\\\ 400=I(1-r)^{\frac{0}{p}}\implies 400=I\cdot 1\implies 400=I[/tex]
therefore then
[tex]\bf \textit{Periodic Exponential Decay}\\\\ A=I(1 - r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &400\\ r=rate\to 20\%\to \frac{20}{100}\to &0.20\\ t=\textit{elapsed time}\\ p=period\to &2 \end{cases} \\\\\\ A=400(1-0.2)^{\frac{t}{2}}[/tex]
now, how many units will it had decreased by the 9th year? t = 9
[tex]\bf A=400(1-0.2)^{\frac{9}{2}}[/tex]
146.54 units are the total units will she sell her ninth year.
Here, we have,
periodic exponential decay is:
[tex]A = I(1-r)^{t/p}[/tex]
where,
A = amount
I = initial amount
r = rate
t = time
p = period
we know, the first year she sold 400 units,
thus year 0, 400 units, t = 0, A = 400.
so, we get,
periodic exponential decay is:
[tex]A = I(1-r)^{t/p}[/tex]
=> 400 = I × 1
=> 400 = I
therefore then, when, rate = 20 %
we get,
periodic exponential decay is:
[tex]A = I(1-r)^{t/p}[/tex]
[tex]= > A = 400(1-0.2)^{t/2}[/tex]
now, how many units will it had decreased by the 9th year? t = 9
so, we get,
[tex]= > A = 400(1-0.2)^{9/2}[/tex]
=> A = 146.54
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Determine whether the function f : z × z → z is onto if
a.f(m,n)=m+n. b)f(m,n)=m2+n2.
c.f(m,n)=m.
d.f(m,n) = |n|.
e.f(m,n)=m−n.
Answers
b. No; [tex]m^2+n^2\ge0[/tex] for any integers [tex]m,n[/tex]
c. Yes; self-evident
d. No; similar to (b), because [tex]|n|\ge0[/tex] for any [tex]n\in\mathbb Z[/tex]
e. Yes; [tex]\mathbb Z[/tex] is closed under subtraction
The five-number summary for scores on a statistics test is 11, 35, 61, 70, 79. in all, 380 students took the test. about how many scored between 35 and 61
Answers
Answer: There are 95 students who scored between 35 and 61.
Step-by-step explanation:
Since we have given that
The following data : 11,35,61,70,79.
So, the median of this data would be = 61
First two data belongs to "First Quartile " i.e. Q₁
and the second quartile is the median i.e. 61.
The last two quartile belongs to "Third Quartile" i.e. Q₃
And we know that each quartile is the 25th percentile.
And we need "Number of students who scored between 35 and 61."
So, between 35 and 61 is 25% of total number of students.
So, Number of students who scored between 35 and 61 is given by
[tex]\dfrac{25}{100}\times 380\\\\=\dfrac{1}{4}\times 380\\\\=95[/tex]
Hence, There are 95 students who scored between 35 and 61.
The number of students who scored between 35 and 61 is 95
The 5 number summary is the value of the ;
Minimum Lower quartile Median Upper quartile and Maximum values of a distribution.The total Number of students = 380
The lower quartile (Lower 25%) = 35
The median (50%) = 61
The Number of students who scored between 35 and 61 : 50% - 25% = 25%This means that 25% of the total students scored between 35 and 61.
25% of 380 = 0.25 × 380 = 95Hence, 95 students scored between 35 and 61.
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What is the answer to the problem below?
Solve the system of equations graphically.
x=3
y=4
Answers
They already give us the answer, which is (3,4). So all we have to do is graph it on the graph. 3 is for x and 4 is for y.
Check the picture below.
(3,4)
You could solve this graphically by graphing the vertical line x = 3 and the horiz. line y = 4. A visual check would show that these two lines intersect at (3,4).
What is the unit rate for 822.6 km in 18 hours? Enter your answer, as a decimal, in the box. Please Help
Answers
answer
unit rate = 45.7 km per hour
822.6 km
-------------- = 45.7 km/hr
18 hrs
Note: this is approx 45.7 km/hr 0.625 mile
--------------- * ----------------- = 28.6 mph
1 1 km
Forty percent of the players on a soccer team are experienced players. The coach wants to increase this percent. There are 10 players on the team now. The expression below represents the percent of experienced players on the team after the coach adds x experienced players.
100(x+4)/x+10
What is:
a. Original ratio of experienced to total players
b. Final ratio of experienced to total players
c. Final total number of players
Choices
4/10
x+10
4
10
x+4/x+10
Answers
Given that forty percent of the players on a soccer team are experienced players.
Then, the original ratio of experienced to total players is given by 40/100 = 4/10
Part B:
Given that the expression [tex] \frac{100(x+4)}{x+10} [/tex] represents the percent of experienced players on the team after the coach adds x experienced players.
Then, the final ratio of experienced to total players is given by:
[tex] \frac{\frac{100(x+4)}{x+10}}{100} =\frac{x+4}{x+10}[/tex]
Part C:
Given that there are 10 players on the team now, and that the coach adds x experienced player, then the number of players on the team now is given by:
10 + x.
What is the completely factored form of x3 – 64x? x(x – 8)(x – 8) (x-4)(x2+4x+16) x(x – 8)(x + 8) (x – 4)(x + 4)(x + 4)
Answers
After factoring x from both terms, you can factor the difference of two squares.
= x(x² -64)
= x(x -8)(x +8)
_____
It is worth remembering the "speciall form" that is the difference of two squares:
... a² - b² = (a -b)(a +b)
Answer:
x(x -8)(x +8)
Step-by-step explanation: