If the figure below were reflected across the waxis, what would be the new coordinates of point A
Answers
The coordinates of point A are (-2,3). A reflection across the y-axis is given by:
[tex](x,y)\rightarrow(-x,y)[/tex]Applying this rule to point A we have:
[tex](-2,3)\rightarrow(2,3)[/tex]Therefore, the image of point A is (2,3) and the correct option is B.
Related Questions
A triangle on a coordinate plane is translated according to the rule T-3,5(X,Y) what is another way to write this ?
Answers
Given the translation rule as :
[tex]T_{-3,5}(x,y)[/tex]Solution
Another way of writing this is:
[tex](x,\text{ y) }\rightarrow\text{ (}x\text{ - 3, y + 5)}[/tex]This means that the original coordinates (x,y) would be translated 3 units to the left and 5 units upwards to give the new coordinates.
Answer: Option A
I need help with this. Also, i’m aware you can’t see all the graphs listed so just let me know what coordinates would be appropriate and i’ll choose whichever graph has those coordinates.
Answers
Answer: Provided the sunglasses inventory which has the number of sunglasses and days in two columns, we have to find the graph which represents this table.
The table can be modeled by a linear equation:
[tex]\begin{gathered} y(x)=mx+b\Rightarrow(1)\Rightarrow\text{ y glasses as function of days x} \\ \\ \end{gathered}[/tex]Finding the slope and y intercept of this equation (1) leads to the following:
[tex]\begin{gathered} m=\frac{\Delta y}{\Delta x}=\frac{(42-58)}{(10-2)}=\frac{-16}{8}=--2 \\ \\ \\ \\ y(x)=-2x+b \\ \\ 58=-2(2)+b\Rightarrow b=58+4=62 \\ \\ \\ \therefore\Rightarrow \\ \\ y(x)=-2x+62\Rightarrow(2) \end{gathered}[/tex]The answer, therefore, is the plot of equation (2) which is as follows:
If twice the age of a stamp is added to the age of a coin, the result is 45. The difference between three times the age of a stamp and the age of a coin is 5. What is the age of the stamp?
Answers
10 years
1) Considering that we can call the age of a stamp by "s" and the age of a coin by "c" we can write out the following system of Linear Equations:
[tex]\begin{gathered} 2s+c=45 \\ 3s-c=5 \end{gathered}[/tex]Note that we can solve it using the Elimination Method.
2) So let's add simultaneously both equations:
[tex]\begin{gathered} 2s+c=45 \\ 3s-c=5 \\ -------- \\ 5s=50 \\ \frac{5s}{5}=\frac{50}{5} \\ s=10 \end{gathered}[/tex]We can plug into that s=10 and find the age of a coin as well:
[tex]\begin{gathered} 2(10)+c=45 \\ c=45-20 \\ c=20 \end{gathered}[/tex]Note that we subtracted 20 from both sides.
3) Hence, the age of a stamp is 10 years
Find the area of a circle with a circumferenceof 261 feet.
Answers
It is given that Circumference of circle is 26П feet
The expression for the Circumference of circle is 2Пr.
So,
[tex]\begin{gathered} 2\Pi r=26\Pi \\ r=\frac{26\Pi}{2\Pi} \\ r=13 \end{gathered}[/tex]Radius of the circle is 13 feet
The expression for the area of circle is ПRadius²
[tex]\begin{gathered} \text{Area of circle =}\Pi\times r^2^{} \\ \text{Area of circle=3.14}\times13\times13 \\ \text{Area of circle = 530}.93\text{ f}eet^2 \end{gathered}[/tex]Answer: 530.9
I am having a tough time solving this problem from my prep guide, can you explain it to me step by step?
Answers
The range in the average rate of change in temperature of the substance is from a low temperature of -[tex]22^{0}[/tex]F to a high of [tex]16^{0}[/tex]F
The domain of the function f(x) = sin x includes all real numbers, but its range is −1 ≤ sin x ≤ 1. The sine function has different values depending on whether the angle is measured in degrees or radians. The function has a periodicity of 360 degrees or 2π radians.
Given f(x) = -19sin(7/3x + 1/6) – 3
We have to the range in the average rate of change in temperature of the substance is from a low temperature of ___F to a high of ___F
We know that the range of sin x is [-1, 1]
f(x) = -19 sin(7/3x + 1/6) – 3
We know
-1 ≤ sin(7/3x + 1/6) ≤ 1
Now multiply with -19 on both sides
19 ≥ -19sin(7/3x + 1/6) ≤ -19
-19 ≤ -19sin(7/3x + 1/6) ≤ 19
Now subtract 3 from both sides
-19 - 3 ≤ -19sin(7/3x + 1/6) - 3 ≤ 19 - 3
-22 ≤ -19sin(7/3x + 1/6) ≤ 16
-22 ≤ f(x) ≤ 16
Therefore the range in the average rate of change in temperature of the substance is from a low temperature of -220F to a high of 160F
To learn more about the range of sinx functions visit
https://brainly.com/question/18452384
#SPJ9
2x + 2/3y= -2 x, y intercept
Answers
We need to find the points at which the expression below intercept the axis of the coordinate plane:
[tex]2x+\frac{2}{3}y=-2[/tex]To find the "x" intercept we need to find the value of "x" that results in a value of "y" equal to 0. We have:
[tex]\begin{gathered} 2x+\frac{2}{3}\cdot0=-2 \\ 2x+0=-2 \\ 2x=-2 \\ x=\frac{-2}{2}=-1 \end{gathered}[/tex]To find the "y" intercept we need to find which value of "y" the function outputs when we make x equal to 0.
[tex]\begin{gathered} 2\cdot0+\frac{2}{3}y=-2 \\ \frac{2}{3}y=-2 \\ 2y=-6 \\ y=\frac{-6}{2}=-3 \end{gathered}[/tex]The x intercept is -1 and the y intercept is -3.
Nimol talks on the phone [tex]3 \frac{1}{2} [/tex] more than his brother. His parents scolded him and asked him to cut down on phone calls.He reduced[tex] \frac{2}{5} [/tex] of the time he used to. How long did his brother spend talking on the Phone.
Answers
His brother spent talking on the phone
Step - by - Step Explanation
What to find?
Time Nimol's brother spent talking on phone.
Let x be the time Nimol spent in talking on phone.
Let y be the time Nimol's brother spent talking on phone.
x = y + 3 1/2
x =2/5 ( y + 3 1/2)
Solve. 15 = 4n- 5solve for nn=
Answers
15 = 4n -5
Add 5 to both sides of the equation
15 +5 = 4n -5 + 5
20 = 4n
Divide both sides by 4:
20/4 = 4n /4
5 = n
n= 5
What are the coordinates of point B on AC such that the ratio of AB to BC is 5 : 6
Answers
We have a segment AC, with the point B lying between A and C.
The ratio AB to BC is 5:6.
The coordinates for A and C are:
A=(2,-6)
C=(-4,2)
We can calculate the coordinates of B for each axis, using the ratio of 5:6.
[tex]\begin{gathered} \frac{x_a-x_b}{x_b-x_c}=\frac{2-x_b}{x_b+4}=\frac{5}{6}_{} \\ 6\cdot(2-x_b)=5\cdot(x_b+4) \\ 12-6x_b=5x_b+20 \\ -6x_b-5x_b=20-12_{} \\ -11x_b=8 \\ x_b=-\frac{8}{11}\approx-0.72\ldots \end{gathered}[/tex]We can do the same for the y-coordinates:
[tex]\begin{gathered} \frac{y_a-y_b}{y_b-y_c}=\frac{-6-y_b}{y_b-2}=\frac{5}{6} \\ 6(-6-y_b)=5(y_b-2) \\ -36-6y_b=5y_b-10 \\ -6y_b-5y_b=-10+36 \\ -11y_b=26 \\ y_b=-\frac{26}{11}\approx-2.36\ldots \end{gathered}[/tex]The coordinates of B are (-8/11, -26/11).
Amy bought a car in 2009 valued at $32,500. The car is expected to depreciate at a rateof 11.1% annually. In how many years will Amy's vehicle be worth 50% of its originalvalue? Round your answer to the nearest tenth of a year,
Answers
ANSWER :
5.9 years
EXPLANATION :
Exponential function can be expressed as :
[tex]y=A(1\pm r)^t[/tex]where A = initial amount
r = (+) growth or (-) decay rate
t = time
y = amount after t years
From the problem, the initial value of the car is A = $32,500
It depreciates at a rate of 11.1% annually, so r = -11.1% or -0.111
The value of the car will be 50% of its original value, so y = 0.50(32,500) = $16,250
Using the formula above :
[tex]\begin{gathered} 16250=32500(1-0.111)^t \\ \frac{16250}{32500}=(0.889)^t \\ \\ 0.5=(0.889)^t \\ \text{ Take the ln of both sides :} \\ \ln(0.5)=\ln(0.889)^t \\ \ln(0.5)=t\ln(0.889) \\ \\ t=\frac{\ln0.5}{\ln0.889}=5.89\sim5.9yrs \end{gathered}[/tex]If y=kx, where k is a constant, and y=24 when x=6, what is the value of y when x=5?A. 6B. 15C. 20D. 23
Answers
First, we will find the value of k
We can do this by sybstituting y=24, x=6 in;
y=kx and then solve for k
24= k(6)
divide both-side of the equation by 6
24/6 = k
4 = k
k=4
Then when x = 5, we will substitute x=5 and k=4 in; y=kx and then solve for y
y= (4)(5)
y = 20
i really need help writting the slope intercept form
Answers
Equation in slope intercept form is written as
y = mx + b
If slope m = 1/3 and y-intecept b = 3
Equation form using the information above is
[tex]y\text{ =}\frac{1}{3}x\text{ + 3}[/tex]Point slope form using the point (3, 4)
simply use the formula
y - y₁ = m( x- x₁ )
[tex]y\text{ -4=}\frac{1}{3}(x-3)[/tex]
What is the horizontal and vertical shift for the absolute value function below?f(x) =|x-5|+1The graph shifts right 5 and up 1.The graph shifts left 5 and up 1.The graph shifts left 5 and down 1.The graph shifts right 5 and down 1.
Answers
The correct answer is option A;
The graph shifts right 5 and up 1
Use the ordered pairs (3,56) and (7,85) to find the equation of a line that approximates the data. Express your answer in slope-intercept form. If necessary round the slope to the nearest hundredth and the y intercept to the nearest whole number
Answers
Equation of a line in slope-intercept form:
[tex]\begin{gathered} y=mx+b \\ \\ m\colon\text{slope} \\ b\colon y-\text{intercept} \end{gathered}[/tex]1. Find the slope: Use two ordered pairs (x,y) in the next formula:
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \\ \text{Ordered pairs (3,56) and (7,85)} \\ m=\frac{85-56}{7-3}=\frac{29}{4}=7.25 \end{gathered}[/tex]Slope: m=7.25
2. Find the y-interept: Use one ordered pair and the slope to find b:
[tex]\begin{gathered} \text{ordered pair: (3,56)} \\ x=3 \\ y=56 \\ \\ \text{Slope: m=7.25}_{} \\ \\ y=mx+b \\ 56=7.25(3)+b \\ 56=21.75+b \\ 56-21.75=b \\ \\ b=34.25 \\ \\ b\approx34 \end{gathered}[/tex]y-intercept: b= 34
Then, the equation of the line is:[tex]y=7.25x+34[/tex]Net force = ?Net force = ?16 NThe net force for example A isNAThe net force for example B is NA
Answers
Part A
The net force is 4 N up
Part B
The net force is 3 N to the left
80 students scores recorded 68 84 75 82 68 90 62 88 76 93 73 88 73 58 93 71 59 58 5161 65 75 87 74 62 95 78 63 72 66 96 79 65 74 77 95 85 78 8671 78 78 62 80 67 69 83 76 62 71 75 82 89 67 58 73 74 73 6581 76 72 75 92 97 57 63 83 81 82 53 85 94 52 78 88 77 71mean exam score
Answers
Solution
We have the following values:
68,84,75,82,68,90,62,88,76,93,73,79,88,73,58,93,71,59,
58,51,61,65,75,87,74,62,95,78,63,72,66,96,79,65,74,77,95,
85,78,86,71,78,78,62,80,67,69,83,76,62,71,75,82,89,67,58,
73,74,73,65,81,76,72,75,92,97,57,63,68,83,81,82,53,85,94,
52,78,88,77,71
Part a
Range = Max- Min= 97-51= 46
Part b
The mean is given by:
[tex]\text{Mean}=\frac{\sum ^n_{i\mathop=1}x_i}{n}=75[/tex]Part c
The median is given by:
Position 40 ordered= 75 and Position 41 ordered= 75
Then the median is:
[tex]\text{Median}=\frac{75+75}{2}=75[/tex]Part d
The most is the most frequent value and for this case is:
Repeated 5 times
Mode = 78
Part e
The data within the interval 50-54 is:
51 52 53
The variance is given by:
[tex]s^2=\frac{\sum ^n_{i\mathop=1}(x_i-Mean)^2}{n-1}=1[/tex]And the deviation si:
[tex]s=\sqrt[]{1}=1[/tex]
Anthony has already taken 1 quiz during past quarters, and he expects to have 5 quizzes during each week of this quarter. How many weeks of school will Anthony have to attend this quarter before he w have taken a total of 31 quizzes?
Answers
The first step to solve the problem is to create a function that relates the number of quizzes he attends by the number of weeks that elapses. Since he alread took one quizz, then the function must start from that and must grow at a rate of 5 quizzes per week. We have:
[tex]\text{quizzes(w)}=5\cdot w+1[/tex]We want to know how many weeks until he takes 31 quizzes, then we need to make the expression equal to 31 and solve for the value of w. We have:
[tex]\begin{gathered} 5\cdot w+1=31 \\ \end{gathered}[/tex]Then we subtract both sides by 1.
[tex]\begin{gathered} 5\cdot w+1-1=31-1 \\ 5\cdot w=30 \end{gathered}[/tex]Then we divide both sides by 5.
[tex]\begin{gathered} \frac{5\cdot w}{5}=\frac{30}{5} \\ w=6 \end{gathered}[/tex]It'll take 6 weekes before he have taken a total of 31 quizzes.
A training field is formed by joining a rectangle and two semicircle. The rectangle is 87m long and 64m wide. What is the length of a training track running around the field? ( Use the value 3.14 pie, and do not round your answer. Be sure to include the correct unit in your answer.)
Answers
The legth of running around this track will be the length of both semi-circles plus the two bigger sidesof the rectangle.
The length of a semi -circle is half the length of a circle, and can be expressed as:
[tex]S=\frac{2\pi r}{2}=\pi r[/tex]Where r is the radius of the semi-circle. The diameter of the semi-circles is the same as the smaller sides of the rectangle and its radius is half the diameter, so:
[tex]r=\frac{d}{2}=\frac{64}{2}=32[/tex]So, the total length, as said above, is the sum of the length of both semi-circles plus two times the bigger side:
[tex]\begin{gathered} L=2S+2w \\ L=2\cdot\pi r+2\cdot87 \\ L=2\cdot3.14\cdot32+2\cdot87 \\ L=200.96+174 \\ L=374.96 \end{gathered}[/tex]Al the measures used were in meters and length is given in meters to the first power, so the unit of the result is also meters, or just "m".
Thus the answer is 374.96 m.
Solve the following system of equations using the elimination method. Note that the method of elimination may be referred to as the addition method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, enter INFINITELY MANY.)20x − 5y = 208x − 2y = 8(x, y) =
Answers
Given
The system of equations,
20x − 5y = 20
8x − 2y = 8
To find: The solution.
Explanation:
It is given that,
20x − 5y = 20 _____(1)
8x − 2y = 8 _____(2)
That implies,
Divide (1) by 5 and (2) by 2.
Then, (1) and (2) becomes,
4x - y = 4.
Hence, there is infinitely many solution.
Can someone please help me solve #6 on this packet?
Answers
The distance between the two camper stations are 60.44 km and 62.95 km. as calculated using the law of sines.
Let us consider the the first ranger station is A and the second ranger station is C and the camper is at the position B.
It is given that AC = 10 km
∠BAC = 100°
∠BCA = 71°
∴∠ABC = 180 - (100 + 71) = 9
Now we will use this to find the distance between each ranger station and the camper by using the law of sines.
From the law of sines we know that :
[tex]{\displaystyle {\frac {a}{\sin {\alpha }}}\,=\,{\frac {b}{\sin {\beta }}}\,=\,{\frac {c}{\sin {\gamma }}}\,}[/tex]
Now we will use this ratio to calculate the other sides of the triangle.
10 / sin 9 = BC / sin 100
or, BC = 10 × sin 100 / sin 9
or, BC = 62.95 km
Again:
10 / sin 9 = AB / sin 71
or. AB = 60.44 km
Therefore the distance between the two camper stations are 60.44 km and 62.95 km.
To learn more about law of sines visit:
https://brainly.com/question/17289163
#SPJ9
what is 3(x+5) 12 please help I’ve been stuck on it
Answers
Given data:
The given inequality is 3(x+5) >12.
The given inequality can be written as,
[tex]\begin{gathered} 3\mleft(x+5\mright)>12 \\ 3x+15>12 \\ 3x>-3 \\ x>-1 \\ x\in(-1,\text{ }\infty) \end{gathered}[/tex]The graph of the above solution is,
Thus, the solution of the given inequality is (-1, ∞).
Need immediate help on 2 questions for my test tomorrow
Answers
The data that can be determined from the box plots are
a)
This is because the line in the middle of the box plot is the median of the box plot.
b)
The upper quartile of a box plot is the part of the box plot that is to the right of the line. In this case the median line is at the same location and the upper quartile ends at the same location.
and
The data that can not be determined is
c)
The reason this can not be determined is because the median is the average of the grades. This could mean some students scored in the higher levels and more scored below the median, which in turn drags it down.
d)
Would be following the same reason as C. It can be determined that there was a larger range lower grades in period 5 over period 3, but it can't be determined how many.
Leslie has 3 pounds of peanuts, she uses 1 7/8 pounds to make trail mix. How many pounds does she have left?
Answers
What Leslie has is 3 pounds of Peanuts and she has used a fraction of it. That fraction is 1 7/8. To find out how many pounds more she has left, its a matter of subtracting 1 7/8 from 3 whole.
Let the leftover be called x, and you now have;
x = 3 - 1 7/8
Converting the other number to an improper fraction you now have
x = 3 - 15/8
x = 3/1 - 15/8
By using the LCM of both denominators which is 8, the expression now becomes,
x = 24/8 - 15/8
x = 9/8
x = 1 1/8
Les
factoring quadratics h^2+12h+11
Answers
A carnival ride is in the shape of a wheel with a radius of 25 feet. The wheel has 20 cars attached to the center of the wheel. Use 3.14 for pi and round answers to the nearest hundredth, if applicable.a.) What is the measure of each central angle between any two cars? (4 points)b.) What is the arc length of each sector between any two cars? (4 points)c.) What is the area of each sector between any two cars?
Answers
The carnival ride is in shape of wheel with 25ft radius.
The wheel has 20 cars attached to the center of the wheel. Since the cars are evenly distributed, we can thus find the
the measure of the angle between each car by dividing 360 degrees by 20.
#A:
The measure of each central angle between any two cars is:
[tex]\frac{360}{20}=18^0[/tex]#B:
Hence, we can find the length of the arch between any two cars is given by the length of arc formula given below:
[tex]\begin{gathered} \frac{\theta}{360}\times2\pi r \\ \text{where,} \\ r=\text{radius} \\ \theta=\text{measure of each central angle betw}een\text{ two cars} \end{gathered}[/tex]Let us calculate this length below:
[tex]\begin{gathered} \theta=18^0 \\ \frac{18}{360}\times2\pi\times25 \\ =2.5\pi=7.85\text{ (to the nearest hundredth)} \end{gathered}[/tex]#C:
We are asked to find the area of each sector between two cars.
The area of a sector of a circle is:
[tex]\frac{\theta}{360}\times\pi\times r^2[/tex]Since we have all the parameters, let us calculate this area:
[tex]\begin{gathered} Area=\frac{18}{360}\times\pi\times25^2 \\ \\ Area=98.13\text{ (to nearest hundredth)} \end{gathered}[/tex]Therefore, the final answers are:
#A: angle = 18 degrees
#B length = 7.85 feet
#C Area = 98.13 squared feet
A surveyor wants to find the height of a tower used to transmit cellular phone calls. He stands 125 feet away from the tower and meandered the angle of elevation to be 40 degrees. How tall is the tower?
Answers
Given
Answer
[tex]\begin{gathered} \tan 40=\frac{h}{125} \\ 0,84\times125=h \\ h=105\text{ ft} \end{gathered}[/tex]height of tower is 105 ft
Which of the following functions have the ordered pair (2, 5) as a solution?4 + x = yy = 2 x7 - x = yx + 3 = y
Answers
Given
The ordered pair (2,5).
To find which of the functions have the ordered pair as a solution.
Explanation:
It is given that,
The ordered pair (2,5).
Then, put x=2, and y=5 in the function x+3=y.
That implies,
[tex]\begin{gathered} 2+3=5 \\ 5=5 \end{gathered}[/tex]Hence, the ordered pair (2,5) is a solution of the function x+3=y.
Also, substitute x=2, y=5 in the function 7-x=y.
That implies,
[tex]\begin{gathered} 7-2=5 \\ 5=5 \end{gathered}[/tex]Hence, the ordered pair (2,5) is a solution of the function 7-x=y.
A 76.00 pound flask of mercury costs $150.50. The density of mercury is 13.534 g/cm3.It takes 4.800 in^3 of mercury to make one manometer. Find the price of the mercury used to make 21 manometers by first calculating the cost of mercury for one manometer.What is the price of mercury used to make one manometer?
Answers
Price of one pound of mercury is derived as follows;
[tex]\begin{gathered} Price\text{ of 1lb of merc}=\frac{Cost\text{ of flask}}{Pounds\text{ of merc in the flask}} \\ \text{Price of 1 lb}=\frac{150.50}{76} \\ \text{Price of 1 lb}=1.98 \end{gathered}[/tex]This means 1 pound of mercury costs $1.98
1 pound = 453.6 grams
Therefore;
[tex]\begin{gathered} 1lb=453.6gms \\ 76lb=34,473.6gms \end{gathered}[/tex]The price of 1 gram of mercury would be;
[tex]\begin{gathered} Price\text{ of 1 gram}=\frac{Price\text{ per pound}}{\text{grams in 1 lb}} \\ \text{Price of 1 gram}=\frac{1.98}{453.6} \\ \text{Price of 1 gram}=0.004365 \end{gathered}[/tex]This means 1 gram of mercury costs $0.004365
Note that you have 13.534 grams per cubic centimeter of mercury. Therefore, the price of 1 cubic centimeter of mercury shall be calculated as follows;
[tex]\begin{gathered} \text{Price of 1 cubic cm}=grams\text{ per cubic cm x price of 1 gram} \\ \text{price of 1 cubic cm}=13.534\times0.004365 \\ \text{Price of 1 cubic cm}=0.059 \end{gathered}[/tex]This means 1 cubic centimeter of mercury would cost $0.059
Note also that, 1 cubic inch = 16.387 cubic centimeters. Hence,
[tex]\begin{gathered} Price\text{ of 1 cubic inch}=16.387\text{ cubic cm x }price\text{ of 1 cubic cm} \\ \text{Price of 1 cubic inch}=16.387\times0.059 \\ \text{Price of 1 cubic inch}=0.9668 \end{gathered}[/tex]This means 1 cubic inch costs $0.9668
It takes 4.800 cubic inches to make 1 manometer.
Therefore, the cost of 4.800 cubic inches would be;
[tex]\begin{gathered} Price\text{ of 4.800 cubic inches}=Price\text{ of 1 cubic inch x 4.800 cubic inches} \\ Price\text{ of 4.800 cubic inches}=0.9668\times4.800 \\ \text{Price of 4.800 cubic inches}=4.64 \end{gathered}[/tex]If it costs 4.800 cubic inches to make 1 manometer, then the cost of 1 manometer would be $4.64
Therefore, to make 21 manometers, we would have;
[tex]\begin{gathered} 1\text{ manometer}=4.64 \\ 21\text{ manometers}=21\times4.64 \\ 21\text{ manometers}=97.44 \end{gathered}[/tex]ANSWER:
The price of mercury required to make 21 manometers would be $97.44
Graph the linear function using the slope and the y-intercept.y = 2x + 3CORTUse the graphing tool to graph the linear equatium. Use the slope and y-intercept when drawing the line.Click toenlargegraph
Answers
Answer:
Explanation:
If we have a linear equation of the form
[tex]y=mx+b[/tex]then m = slope and b = y-intercept.
Now in our case, we have
[tex]y=2x+3[/tex]which means that slope = 2 and y-intercept = 3
Therefore, we graph a line that has a slope of 2 and a y-intercept of 3.
A slope of 2 means that for every step you take to the right on a graph, you move 2 steps up to get to a point on the line.
The y-intercept of 3 means that the line passes through the point (0, 3).
Using these two facts about the line, we draw the following line.
From the above plot, we can clearly see that the line has a slope of 2 and a y-intercept of 3 - the same line described by y = 2x + 3.
Find cosθ, cotθ, and secθ, where θ is the angle shown in the figure. Give exact values, not decimal approximations.cosθ=cotθ=secθ=
Answers
First let's find the missing value of the hypotenuse:
[tex]\begin{gathered} c^2=a^2+b^2 \\ a=4 \\ b=5 \\ \Rightarrow c^2=(4)^2+(5)^2=16+25=41 \\ \Rightarrow c=\sqrt[]{41} \\ \end{gathered}[/tex]we have that the hypotenuse equals sqrt(41). Now we can find the values of the trigonometric functions:
[tex]\begin{gathered} \cos (\theta)=\frac{adjacent\text{ side}}{hypotenuse} \\ \Rightarrow\cos (\theta)=\frac{4}{\sqrt[]{41}} \\ \sec (\theta)=\frac{1}{\cos (\theta)} \\ \Rightarrow\sec (\theta)=\frac{1}{\frac{4}{\sqrt[]{41}}}=\frac{\sqrt[]{41}}{4} \\ \tan (\theta)=\frac{opposite\text{ side}}{adjacent\text{ side}} \\ \Rightarrow\tan (\theta)=\frac{5}{4} \\ \cot (\theta)=\frac{1}{\tan (\theta)} \\ \Rightarrow\cot (\theta)=\frac{1}{\frac{5}{4}}=\frac{4}{5} \end{gathered}[/tex]