From the given description, it appears that M, L, and K are collinear and the length of MK is equal to the sum of ML and LK.
To be able to find the length of MK let's first find the value of x using the equation of the sum of the lines.
[tex]\text{ }\bar{\text{ML}}\text{ + }\bar{\text{LK}}\text{ = }\bar{\text{MK}}[/tex]Let's plug in the values given in the description.
[tex]\text{ (8) + (x + 2) = 4x - 2}[/tex][tex]\text{ 8 + x + 2 = 4x - 2}[/tex][tex]\text{ x + 10 = 4x - 2}[/tex][tex]\text{ x - 4x = -2 - 10}[/tex][tex]\text{ -3x = -12}[/tex][tex]\text{ }\frac{\text{-3x}}{-3}\text{ = }\frac{\text{-12}}{-3}[/tex][tex]\text{ x = 4}[/tex]Let's plug in x = 4 in the equation for the length of MK = 4x - 2.
[tex]\text{ }\bar{\text{MK}}\text{ = 4x - 2}[/tex][tex]\text{ }\bar{\text{MK}}\text{ = 4(4) - 2}[/tex][tex]\text{ }\bar{\text{MK}}\text{ = 16 - 2}[/tex][tex]\text{ }\bar{\text{MK}}\text{ = 1}4[/tex]Therefore, the length of MK is 14.
using the pythagorean theorem and finding distances
Scalene
Explanation
we have a triangle with 3 differents lengths and angles, so it is a Scalene triangle
A triangle is scalene if all of its three sides are different.
I hope this helps you
• 21 Theodore inherited two different stocks whose yearly income was $2100. The total appraised value of the stocks was $40,000 and one was paying 4% and one 690 per year. What was the value of each stock? o
hello
the yearly income was $2100
the appraised value = $40,000
one of the stocks pays 4% annually
the other pays $690 yearly
let's find how much the 4% stock pays annually
to do this, let's subtract the income of one of the stocks from the total income. i.e 690 from 2100
[tex]2100-690=1410[/tex]the other stock pays $1410 annually
now we can simply find the value of each stock
[tex]\begin{gathered} 4\text{\% of x gives 1410 annually} \\ \frac{4}{100}=\frac{1410}{x} \\ \text{cross multiply both sides } \\ 4\times x=100\times1410 \\ 4x=141000 \\ \text{divide both sides by 4} \\ \frac{4x}{4}=\frac{141000}{4} \\ x=35250 \end{gathered}[/tex]the value of one of the stock is $35250
we can proceed to find the value of the other stock by subtracting 35250 from 40000 which is the value of the two stock
[tex]40000-35250=4750[/tex]from the calculations above, the value of the stocks is $35250 and $4750
Which property of equality would you use to solve the equation 5m = 12?
We would have to use the division (and/or multiplication) property in order to solve, and that would be:
[tex]5m=12\Rightarrow m=\frac{12}{5}[/tex]what are the solutions of the equation 0 equals x ^ 2 + 3x - 10
The given expression is :
[tex]0=x^2+3x-10[/tex]Factorize the expression :
[tex]\begin{gathered} 0=x^2+3x-10 \\ x^2+3x-10=0 \end{gathered}[/tex]Find the pair of number such that : the product of two numbers are equal = (-10)
and thier summation is equal to 3
i.e. 5 x ( -2) = -10 and 5 + (-2) = 3
So,
[tex]\begin{gathered} x^2+3x-10=0 \\ x^2+5x-2x-10=0 \end{gathered}[/tex]Take x common from the first two terms and (-2) from last two terms :
[tex]\begin{gathered} x^2+5x-2x-10=0 \\ x(x+5)-2(x+5)=0 \\ \text{Now, take (x+5) common :} \\ (x-2)(x+5)\text{ =0} \end{gathered}[/tex]Now equate each factor with zero :
[tex]\begin{gathered} (x-2)(x+5)=0 \\ x-2=0\Rightarrow x=2 \\ x+5=0\Rightarrow x=-5 \end{gathered}[/tex]Answer : C) x = -5, 2
i need the range for city A and city B, and the standard deviation for city A and city B
The range in a data set is the highest value minus the lowest value.
The standard deviation is a quantity expressing by how much the members of a group differ from the mean value for the group.
City A (data set)
[tex]1.00,1.00,1.25,1.50,1.50[/tex]City B (data set)
[tex]0.00,1.00,1.75,1.75,2.25[/tex]City A RangeThe range is 1.50 - 1.00 = 0.50
City B Range
The range is 2.25 - 0.00 = 2.25
Now,
The formula for sample standard deviation is
[tex]s=\sqrt[]{\frac{\sum(x-\bar{x})^2}{n-1}}[/tex]Where
s is the sample standard deviation
x bar - mean of the sample
n is the number of numbers in the data set
Using a standard deviation calculator, let's calculate the standard deviation of both data sets.
City A Standard DeviationThe standard deviation is
[tex]s=\sqrt[]{\frac{(1.00-1.25)^2+\cdots(1.50-1.25)^2}{5-1}}=0.25[/tex]City B Standard DeviationThe standard deviation is
[tex]s=\sqrt[]{\frac{(0.00-1.35)^2+\cdots(2.25-1.35)^2}{5-1}}=0.8768[/tex]From the data calculations, we can see that City B
how do i solve 13-3/2x=37
An apple is launched directly upward at 70 feet per second from a platform 100
feet high. The equation for this apple's height h at time t seconds after launch
is - 16t+70t+100. What is the height of the apple after 5 seconds?
Height of apple:
Answer:
530
Step-by-step explanation:
(16x5)+(70x5)+100=?
80+350+100=530
describe your so how to find the following product using both an algorithm in the diagram 3 * 3/4
To multiply the given numbers, we can use 0.75 instead of 3/4.
Then, we multiply using the standard algorithm.
Notice that the result must have 4 decimal numbers since that's the number of decimals both numbers have at the beginning.
Now, we can use the areas model to show a diagram about the multiplication.
Notice that we've got 3 wholes with 3/4 each, that gives us 2.25 because it's the sum 0.75 + 0.75 + 0.75 = 2.25.
Consider the following rational expression:2 – 2y / 2y - 2Step 1 of 2: Reduce the rational expression to its lowest terms.Answer
Factor out 2 on both numerator and denominator
[tex]\begin{gathered} \frac{2-2y}{2y-2} \\ =\frac{2(1-y)}{2(y-1)} \\ \\ \text{cancel out }2\text{ on both numerator and denominator} \\ =\frac{\cancel{2}(1-y)}{\cancel{2}(y-1)} \\ =\frac{(1-y)}{(y-1)} \\ \\ \text{factor out }-1\text{ on numerator},\text{ and rearrange to cancel out common binomial} \\ =\frac{(1-y)}{(y-1)} \\ =\frac{-1(-1+y)}{(y-1)} \\ =\frac{-1(y-1)}{(y-1)} \\ =\frac{-1\cancel{(y-1)}}{\cancel{(y-1)}} \\ =-1 \\ \\ \text{Therefore,} \\ \frac{2-2y}{2y-2}=-1 \end{gathered}[/tex]Part 2:
Since the given expression is in fraction, we cannot let the denominator equal to zero. Find values of y that makes the denominator by zero
[tex]\begin{gathered} \text{Denominator: }2y-2 \\ \\ \text{Equate to zero} \\ 2y-2=0 \\ 2y-2+2=0+2 \\ 2y\cancel{-2+2}=2 \\ \frac{2y}{2}=\frac{2}{2} \\ y=1 \\ \\ \text{If }y=1,\text{ the denominator }2y-2\text{ becomes zero therefore}, \\ y\neq1 \end{gathered}[/tex](X^-3y^2/x^3)^-2
Simplify the expression. Your final answer should use positive exponents.
Answer:
y^-4
here you are
,........
Complete the table for each function. Then answer the questions that follow.X Y=4x Y=4x^2 Y=4^x0 0 0 11 4 a b2 c 16 d3 e f ga=? b=?c=?d=?e=?f=?g=?
Let's evaluate each function for the integers from 0 to 3:
[tex]undefined[/tex]A table of 5 students has 2 seniors and 3 juniors. The teacher is going to pick 2 students at random from this group to present homework solutions. Find the probability that both students selected are juniors
ANSWER
[tex]\text{ P\lparen both students are junior\rparen = }\frac{1}{10}[/tex]EXPLANATION
Given information
The total number of junior students = 2
The total number of senior students = 3
The total number of students = 5
To determine the probability of picking two junior students, follow the steps below
Step 1: Define probability
[tex]\text{ Probability = }\frac{\text{ possible outcome}}{\text{ total outcome}}[/tex]Step 2: Find the probability of picking the first junior students
[tex]\begin{gathered} \text{ Probability = }\frac{possible\text{ outcome}}{total\text{ outcome}} \\ \text{ Probability of picking the first junior students is} \\ \text{ P\lparen Junior student\rparen = }\frac{2}{5} \end{gathered}[/tex]Assuming the first picking was successful, then, we will be left with 1 junior student and 3 senior students.
Therefore, the new total outcome can be calculated below
1 + 3 = 4 students
Step 3: Find the probability that the second picking will be a junior student
[tex]\begin{gathered} \text{ Probability = }\frac{\text{ possible outcome}}{\text{ total outcome}} \\ \text{ P\lparen picking the second junior student\rparen = }\frac{1}{4} \end{gathered}[/tex]Step 4: Find the probability that both students are junior students
[tex]\begin{gathered} \text{ P\lparen both students are junior students\rparen = }\frac{2}{5}\times\frac{1}{4} \\ \text{ P\lparen both students are junior students\rparen = }\frac{2}{20} \\ \text{ P \lparen both students are junior students \rparen = }\frac{1}{10} \end{gathered}[/tex]Hence, the probability that both students selected are juniors is 1/10
15. Write the equation of a horizontal line going through the point (5,4). Equation:
Given the point: (5, 4)
We know that the general equation of a horizontal line is y = k
Where k is constant.
From the point: (5, 4)
The only way for a line that is horizontal to satisfy (x, y) ==> (5, 4) is if y = 4.
To determine the equation of a horizontal line, the x-corrdinate is ignored because a horizontal line touches all possible values of x.
Therefore, the equation of the horizontal line is:
y = 4
ANSWER:
y = 4
Which choices are equivalent to the expression below? Check all that apply.A.B.C.72D.E.F.
GIven:
[tex]3\sqrt{8}[/tex]Required:
We need to find the equivalent expression
Explanation:
let
[tex]\begin{gathered} x=3\sqrt{8} \\ x^2=72 \end{gathered}[/tex]now just we need to check that which square is 72
1)
[tex]\begin{gathered} a=\sqrt{3}\sqrt{12} \\ a^2=36\text{ not possible} \end{gathered}[/tex]2)
[tex]\begin{gathered} b=\sqrt{6}\sqrt{12} \\ b^2=72\text{ possible} \end{gathered}[/tex]3)
[tex]\begin{gathered} c=72 \\ c^2=5184\text{ not possible} \end{gathered}[/tex]4)
[tex]\begin{gathered} d=\sqrt{3}\sqrt{24} \\ d^2=72\text{ possible} \end{gathered}[/tex]5)
[tex]\begin{gathered} e=\sqrt{6}\sqrt{24} \\ e^2=144\text{ not possible} \end{gathered}[/tex]6)
[tex]\begin{gathered} f=\sqrt{9}\sqrt{8} \\ f^2=72\text{ possible} \end{gathered}[/tex]Final answer:
[tex]\begin{gathered} \sqrt{6}\sqrt{12} \\ \sqrt{3}\sqrt{24} \\ \sqrt{9}\sqrt{8} \end{gathered}[/tex]
are equivalent to given expression
Solve. Show all your work!The digits of a positive two-digit integer N are interchanged to form an integer K. Find allpossibilities for N if N is even and exceeds K by more than 50.
Let the units place digit be U and the tens place digit be T.
The number N is given by:
[tex]N=10T+U\ldots(i)[/tex]The number K is given by:
[tex]K=10U+T\ldots(2)[/tex]It is given that N is even that means U can be only from 0,2,4,6,8.
It is also given that N exceeds K by more than 50 so it follows:
[tex]\begin{gathered} N-K\ge50 \\ 10T+U-(10U+T)\ge50 \\ 9T-9U\ge50 \end{gathered}[/tex]So it can be said that:
[tex]T-U\ge\frac{50}{9}\approx5.5556\approx6[/tex]Since the value of T-U will always be an integer and it should be greater than or equal to 6.
The number T can be 1 to 9 and U can be only 0,2,4,6,8 so it follows:
[tex]\begin{gathered} T=9,U=0\Rightarrow T-U=9 \\ T=9,U=2\Rightarrow T-U=7 \\ T=8,U=0\Rightarrow T-U=8 \\ T=7,U=0\Rightarrow T-U=7 \\ T=6,U=0\Rightarrow T-U=6 \\ T=8,U=2\Rightarrow T-U=6 \end{gathered}[/tex]Hence the possible values for integer N are 90,92,80,70,60,82 and the respective integer K will be 09,29,08,07,06,28.
In all cases the difference is more than 50 as you can check.
Describe the end behavior for the polynomial function described below.
ANSWER
As x → ∞, f(x) → ∞
As x → -∞, f(x) → -∞
EXPLANATION
The degree of this polynomial is 5, which is odd. Therefore, the ends of the graph will point one up and the other down.
The coefficient is positive - the coefficient is the coefficient of the variable with the highest degree, in this case x⁵ - so the right end (when x → ∞) points up (to ∞) and the left end points down (to -∞).
John got all 20 questions on math test all were right plus an additional bonus question correctly. Bonus was worth same as other questions. What was the test score in percent?
The question on the math test is
[tex]=20[/tex]The bonus question is
[tex]=1[/tex]The total questions scored correctly
[tex]\begin{gathered} =20+1 \\ =21 \end{gathered}[/tex]To calculate the test score in percent, we will use the formula below
Since she got all the questions plus the bonus question correctly( note: all questions worth the same points), then we will have
[tex]\text{test score(percent)=}\frac{\text{Total questions scored correctly}}{\text{Total questions on the math test}}\times100\text{ \%}[/tex]By substituting the values, we will have
[tex]\begin{gathered} \text{test score(percent)=}\frac{\text{Total questions scored correctly}}{\text{Total questions on the math test}}\times100\text{ \%} \\ \text{test score(percent)}=\frac{21}{20}\times100 \\ \text{test score(percent)}=\frac{2100}{20} \\ \text{test score(percent)}=105\text{ \%} \end{gathered}[/tex]Hence,
The test score in percent = 105%
What is the mean of the data?Number of Letters inOur First Name3 XX4 XX5 XXXX6 X7 X8 X I keep getting 3 but it was marked wrong with the correct answer of 5. Where am I going wrong?
Answer
Mean = 5
Explanation
The mean is the average of the distribution. It is obtained mathematically as the sum of variables divided by the number of variables.
Mean = (Σx)/N
x = each variable
Σx = Sum of the variables
N = number of variables
Since the number of X in front of each number denotes how many times that number occurs,
Σx = 3 + 3 + 4 + 4 + 5 + 5 + 5 + 5 + 6 + 7 + 8 = 55
N = Total number of X = 11
Mean = (Σx)/N
Mean = (55/11) = 5
Hope this Helps!!!
Creating and solving equationstwo-thirds a number plus 4 is 7
Given:
two-thirds a number plus 4 is 7
First Part: Converting the statement into equation
Let x be the number in the given statement.
The phrase "two-thirds a number" can be expressed as
[tex]\frac{2}{3}x[/tex]Pair it with "... plus 4" and we get
[tex]\frac{2}{3}x+4[/tex]Finally, it is stated it is equal to 7, and we complete the equation
[tex]\frac{2}{3}x+4=7[/tex]Second Part: Solving for the number
Now, that we have the equation, we can now solve for the missing number x.
Subtract both sides by 4, to remove the constant 4 on the left side of the equation
[tex]\begin{gathered} \frac{2}{3}x+4=7 \\ \frac{2}{3}x+4-4=7-4 \\ \frac{2}{3}x\cancel{+4-4}=3 \\ \frac{2}{3}x=3 \end{gathered}[/tex]Multiply both sides by 3/2, and we get
[tex]\begin{gathered} \frac{2}{3}x=3 \\ \frac{2}{3}x\cdot\frac{3}{2}=3\cdot\frac{3}{2} \\ \frac{\cancel{2}}{\cancel{3}}x\cdot\frac{\cancel{3}}{\cancel{2}}=\frac{9}{2} \\ x=\frac{9}{2} \end{gathered}[/tex]Therefore, the number is 9/2 or nine-halves.
the perimeter of the original rectangle is 16 ft. what is the perimeter of the enlarged rectangle? round to the nearest tenth if necessary.
To calculate the perimeter of the enlarged rectangle, we need to get the breadth first.
The original rectangle is similar to the enlarged rectangle, so the ratio of thier corresponding sides must be equal
Let b represent the breadth of the enlarged rectangle,
[tex]\begin{gathered} \text{ Ratio of corresponding sides is given as } \\ \frac{1.8}{b}=\frac{6.2}{12.4} \\ \text{cross multiply} \\ 6.2b=12.4\text{ x 1.8} \\ 6.2b=22.32 \\ b=\frac{22.32}{6.2} \\ b=3.6\text{ ft} \end{gathered}[/tex]Now, the length of the enlarged rectangle is 12.4ft while the breadth is 3.6ft
The perimeter = 2(L+B)
= 2(12.4 + 3.6)
=2(16)
=32ft
The perimeter of the enlarged rectangle is 32 ft
A carpenter cuts a 5-ft board in two pieces. One piece must be three times as longas the other. Find the length of each piece.
3.75 ft and 1.25 ft
Explanation
Step 1
Diagram
Step 2
set the equations
let x represents the longest piece
lety represents the smaller piece
so
a)A carpenter cuts a 5-ft board in two pieces, hence
[tex]x+y=5\Rightarrow equation(1)[/tex]b)One piece must be three times as long as the other,then
[tex]x=3y\Rightarrow equation(2)[/tex]Step 3
finally, solve the equations:
a) replace the x value from equation (2) into equation(1)
[tex]\begin{gathered} x+y=5\Rightarrow equation(1) \\ (3y)+y=5 \\ add\text{ like terms} \\ 4y=5 \\ divide\text{ both sides by 4} \\ \frac{4y}{4}=\frac{5}{4} \\ y=1.25 \end{gathered}[/tex]b) now, replace the y value into equation (2) to find x
[tex]\begin{gathered} x=3y\Rightarrow equation(2) \\ x=3(1.25) \\ x=3.75 \end{gathered}[/tex]therefore, the lengths of the pieces are
3.75 ft and 1.25 ft
I hope this helps you
The points ( 0.5 , 1/10 ) and ( 7 , 1 2/5)The points are on the graph of a proportional relationshipIt is required to find the constant of proportionally
Given: the points ( 0.5 , 1/10 ) and ( 7 , 1 2/5)
The points are on the graph of a proportional relationship
It is required to find the constant of proportionally
so, the graph of the points will be as shown in the following image:
as shown there a proportional relationship
Because the line is pass through zero
So, the constant of proportionality = y/x
It can be calculated using any point from the given points
So ,
Using the point ( 0.5 , 1/10)
the constant = 0.1/0.5 = 0.2
We can check the answer using the other point ( 7, 1 2/5)
The constant = (1 2/5)/7 = 1.4/7 = 0.2
So, the constant of proportionality = 0.2
Suppose a Ballon rises 250 ft during the first minute.................
Since for each succeeding minute the balloon rises 70% as much as it did in the previous minute, and in the first minute the balloon rises 250 ft, then in the minute n the balloon rises
[tex]250*0.70^{n-1}ft.[/tex]Therefore the height of the balloon in the minute n is
[tex]\sum_{k\mathop{=}1}^n250*0.70^{k-1}ft.[/tex]Therefore the height of the balloon in the minute 8 is:
[tex]\begin{gathered} 250*0.70^0ft+250*0.70^1ft+250*0.70^2ft+250*0.70^3ft \\ +250*0.70^4ft+250*0.70^5ft+250*0.70^6ft+250*0.70^7ft \\ \approx785.30ft \end{gathered}[/tex]Answer: Option d.
a) What were the ranges of typing speeds for the two groups?Group 1: Group 2:b) Which group had more typing speed in the 40s1) Group 1 2)Group 2 3) Each had the samec) Which group had the greater median typing speed?1) Group 1 2)Group 2 3) Each had the same
The given stem and leaf plot shows the typing speeds of two groups of students.
Group 1 has n1= 20 students
Group 2 has n2= 19 students
The stem and leaf plot is two sided, meaning that they share the same stem.
The observed values are the number of words per minute.
In the steam the ten of each value is placed and in the leafs you find the units:
This way you can determine the observations for both samples. I'll do so and arrange them form least to greatest:
Group 1:
33, 34, 37, 42, 44, 44, 45, 47, 48, 49, 49, 50, 51, 52, 52, 55, 55, 63, 66, 67
Group 2:
33, 36, 41, 42, 44, 46, 46, 51, 52, 52, 52, 53, 53, 56, 59, 60, 66, 67, 69
Part a
The range is calculated as the difference between the maximum and minimum observations of a sample. To determine those values you need the sample ordered from least to greatest.
For group 1:
Minimum value: 33 words/min
Maximum value: 67 words/min
Range= maximum-minimum=67-33=34words/min
For group 2:
Minimum value: 33words/min
Maximum value: 69 words/min
Range: 69-33=36words/min
→ the range for group 1 is 34words/min while the range for group 2 is 36words/min
Part b
To determine which group had more typing speeds in the fourties you have to count said observations for both of them.
You can do it directly from the stem and leaf plot, go to the row correpsonding to the 4 in the plot and count or use the values:
For group 1: in the second row there are 8 leafs, corresponding to the observations: 42, 44, 44, 45, 47, 48, 49, 49,
For group 2: in the second row there are 5 leafs, corresponding to the observations: 41, 42, 44, 46, 46
→There are more typing speeds in the 40s in group 1.
Part c:
The median is a measure of center that divides the sample in two halves. To calculate it you have to determine its position and then look for the corresponding value in the sample that was previously ordered from least to greatest.
To determine the position of the mean you have to use the following fomula:
For even samples: n/2
For odd samples: (n+1)/2
Median of group 1
n1=20 students
The sample is even, calculate its position using the first formula:
Position: n/2 = 20/2= 10
The median is in the tenth position, look in the sample for the tenth observation:
33, 34, 37, 42, 44, 44, 45, 47, 48, 49, 49, 50, 51, 52, 52, 55, 55, 63, 66, 67
→The median typing speed for group 1 is 49 words/min
Median of group 2
n2=19 students
The sample is odd, you have to use the second formula to find its position:
Position: (n+1)/2= (19+1)/2= 20/2= 10
The median of this group is the 10th observation:
33, 36, 41, 42, 44, 46, 46, 51, 52, 52, 52, 53, 53, 56, 59, 60, 66, 67, 69
→The median typing speed for group 2 is 52 words/min
→Group 2 has the greater median typing speed.
The sum of two numbers is 16. Their diference is 20. Find the numbers
Step 1
Let the numbers be x and y and let us assume x is the bigger number
The equations then are;
[tex]\begin{gathered} x+y=16----(1) \\ x-y=20--(2) \end{gathered}[/tex]Step 2
Add equations 1 and 2
Hence,
[tex]\begin{gathered} 2x=36 \\ \frac{2x}{2}=\frac{36}{2} \\ x=18 \end{gathered}[/tex]Step 3
Find the value of y from equation 2
[tex]undefined[/tex]Consider the polynomial function p given by p(x) = 728 – 2x2 + 32 + 10. Evaluate the functionat x = -3
Answer:
-206
Step-by-step explanation:
We are given the following function:
p(x) = 7x³ - 2x² + 3x + 10
To evaluate it at x = -3, we replace x by -3. So
p(-3) = 7x³ - 2x² + 3x + 10 = 7*(-3)³ - 2*(-3)² + 3*(-3) + 10 = 7*(-27) - 2*9 - 9 + 10 = -189 - 18 + 1 = -189 - 17 = -206
identify the horizontal asymptotes, if they exist, for the following function…
You have the following function:
[tex]f(x)=\frac{5x^4-2x}{x^4+32}[/tex]Take into account that if the expression for the numerator has the same degree that the expression at the denominator, the horizontal asymptote is given by the quotient between the leading coefficient of each polynomial.
In this case, leading coeffcicient of numerator is 5 and from the denominator we get a leading coeeficient equal to 1.
Then, the horizontal asymptote is:
y = 5/1 = 5
1. P, Q and R are three buildings. A car began its journey at P, drove to Q, then to R and returned to P. The bearing of Q from P is 058º and R is due east of Q. PQ = 114 km and QR = 70 km. © Draw a clearly labelled diagram to represent the above informationen on the diagram TƏRund (a) the north/south direction (b) the bearing 058° (c) the distances 114 km and 70 km. (ii) Calculate (a) the measure of angle POR (b) the distance PR [3] (c) the bearing of P from R [3]
Step 1
Given;
[tex]\begin{gathered} The\text{ bearing of Q from P is 058}^o\text{ } \\ R\text{ is due east of Q} \\ PQ=114km \\ QR=70km \end{gathered}[/tex]Step 2
Draw the diagram
Step 2
Calculate the measure of angle PQR
[tex]\angle PQR=58+90=148^o[/tex]This is because using alternate exterior angles are equal theorem, the first part of angle Q 58 degrees. Since R is due east of Q, then the other part must be 90 degrees, when summed we get 148 degrees
Step 3
Calculate the distance PR. To do this we will use the cosine rule
[tex]\begin{gathered} PR^2=PQ^2+QR^2-2PQ\left(QR\right?cosQ \\ PR^2=114^2+70^2-2\left(114\right)\left(70\right)cos\left(148\right) \\ PR^2=17896+13534.84761 \\ PR=\sqrt{31430.84761} \\ PR=177.2874717 \\ PR\approx177.3km\text{ to the nearest tenth} \end{gathered}[/tex]Step 4
Calculate the bearing of P from R.
Use sine rule and find angle R
[tex]\frac{sin\text{ 148}}{177.2874717}=\frac{sinR}{114}[/tex][tex]\begin{gathered} 114sin148=177.2874717sinR \\ R=\sin^{-1}\frac{\mleft(114sin148\mright)}{177.2874717} \\ R=19.92260569 \end{gathered}[/tex]The bearing of P from R = (90-angle R)+90+90=250 degrees approximately to the nearest whole number
[tex]\begin{gathered} =\left(90-19.92260569\right)+90+90 \\ =250.07739 \\ \approx250^o \end{gathered}[/tex]The bearing of P from R =250 degrees approximately to the nearest whole number
For the data in the table, does y very direction with x? If it does, write an equation for direct variation. Show your work. (2 points)
Hint: y=kx is the formula for direct variation. Find k, then substitute it back in to the formula.
In the given table, the values of y vary directly with the value of x. The equation for direct variation is y = 2.5x
Does y vary directly with x?Direct variation is when two variables move in the same direction. If one variable increases, the other variable increases.
The equation that is used to represent direct variation is y = kx
Where:
y = dependent variable k = constant of variation x = independent variableIf y and x have a direct variation, the constant of variation for all the values of x and y would be equal.
k = y / x
5 / 2 = 2.5
15 / 6 = 2.5
25 / 10 = 2.5
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On a piece of paper, graph y+25**-1. Then determine which answer choicematches the graph you drew.ABСD0.9.-3)0,-)(0-3)69,-2)(4-2)(4.23(2)O A. Graph AB. Graph BO C. Graph CO D. Graph D
Let's graph the given inequality:
As we can see, it matches graph A from the options we were given.