The AAS triangle congruence postulate proves that these triangles are congruent .
In the question,
two triangles are given that are triangle ABC and triangle PQR .
Consider the triangle ABC and triangle PQR .
we can see that
(i) angle C = angle R .....given in the figure
(iii) angle B = angle Q .... given in the figure
(ii) side BC = side QR ....given in the figure
From the above three statements we conclude that
ΔABC ≅ ΔPQR
both the triangles KLM and PQR are congruent by AAS Congruence Postulate .
Therefore , The AAS triangle congruence postulate proves that these triangles are congruent .
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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
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
What is the equation for the linear model in the scatterplot obtained by choosing the two points closest to the line
consider two points closest to the line. say ,
[tex]\begin{gathered} (x_1,y_1)=(6,0) \\ (x_2_{}_{}_{},y_2)=(8,1) \end{gathered}[/tex]let us find the slope, m by the formula
[tex]m=\frac{y_2-y_1}{x_2_{}_{}-x_1}[/tex]subsitute the points in the formula,
[tex]\begin{gathered} m=\frac{1-0}{8-6} \\ m=\frac{1}{2} \end{gathered}[/tex]let us find the y - intercept.
[tex]y=mx+b\ldots(1)[/tex]subsitute the one of the point (6,0) in the above equation.
[tex]\begin{gathered} 0=\frac{1}{2}\times6+b \\ 0=3+b \\ b=-3 \end{gathered}[/tex]thus,
subsitute m= 1/2 and b = - 3 in the equation (1),
[tex]y=\frac{1}{2}x-3[/tex]what digit is in the
Rounding each number to the nearest ten:
• 96 = 100
,• 63 = 60
,• 27 = 30
,• 76 = 80
Sum with rounded numbers:
[tex]100\text{ + 60+30+80=270}[/tex]Answer = 270
how do i solve 13-3/2x=37
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
An account earns an annual rate of 5.4% compounded monthly. If $3,000 is deposited into this account, then after 3 years there is $___. Round your answer to two decimals.
Given:
rate (r) = 5.4% or 0.054 in decimal form
Principal (P) = $3,000
time in years (t) = 3 years
number of conversions per year (m) = 12 (because it says monthly)
Find: future value or maturity value
Solution:
The formula for getting the future value of a compound interest is:
[tex]F=P(1+\frac{r}{m})^{mt}[/tex]Let's plug in the given data to the formula above.
[tex]F=3,000(1+\frac{0.054}{12})^{12\times3}[/tex]Then, solve for F or future value.
[tex]\begin{gathered} F=3,000(1.0045)^{36} \\ F=3,000(1.17532999) \\ F\approx3,526.30 \end{gathered}[/tex]Answer: After 3 years, the deposited money will become $3, 526.30.
Which graph represents the function f(x) = -x + 31?
Answer:
Step-by-step explanation:
I hope this helps! :) If it does could you please mark me brainliest?
Answer:
Slope : -1
y = intercept : (0,31)
Step-by-step explanation:
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
Algebraically determine whether each of the following functions is even, odd or neither. then graph it B. y = x^3 – 3 C. y = 2x^3 - x
According to the even and odd function rules, we found out that the function [tex]y=x^{3}-3[/tex] is neither even nor odd and the function [tex]y=2x^{3}-x[/tex] is an odd function.
It is given to us that the functions are -
B. [tex]y=x^{3}-3[/tex]
C. [tex]y=2x^{3}-x[/tex]
We want to determine each of the following functions is even, odd or neither.
To see if the function is even, we have to check if [tex]f(-x)=f(x)[/tex]
To see if the function is odd, we have to check if [tex]f(-x)=-f(x)[/tex]
B. Here, we have
[tex]y=x^{3}-3\\= > f(x)=x^{3}-3\\= > f(-x)=(-x)^{3}-3\\= > f(-x)=-x^{3}-3[/tex]
We see that [tex]f(-x)\neq f(x)[/tex]. This implies that the function is not even.
Also, [tex]f(-x)\neq -f(x)[/tex]. This implies that the function is not odd.
Therefore, this function is neither even nor odd.
C. Here, we have
[tex]y=2x^{3}-x\\= > f(x)=2x^{3}-x\\= > f(-x)=2(-x)^{3}-(-x)\\= > f(-x)=-2x^{3}+x[/tex]
We see that [tex]f(-x)\neq f(x)[/tex]. This implies that the function is not even.
However,
[tex]f(-x)=-2x^{3}+x\\= > f(-x)= -(2x^{3}-x)\\ = > f(-x)=-f(x)[/tex]
This implies that the function is odd.
Therefore, this function is odd.
Thus, the function [tex]y=x^{3}-3[/tex] is neither even nor odd and the function [tex]y=2x^{3}-x[/tex] is an odd function.
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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
Hi!A particle moves along a straight line, so its speed is () = ^2 − + 6, where t is the time measured in seconds and the speed is measured in meters timessecond.a) Calculate the distance traveled between the seconds t=1 and t=3
In this problem
the distance traveled between the seconds t=1 and t=3 is given by
[tex]\int_1^3(t^2-t+6)dt=\frac{50}{3}\text{ m}[/tex]The answer is
50/3 metersor 16.67 metersExplanation of integrals
In this problem we have
[tex]\int_1^3(t^2-t+6)dt=\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt[/tex][tex]\begin{gathered} \int_1^3t^2dt=\frac{t^3}{3} \\ Evaluate\text{ at 3 and 1} \\ \frac{(3)^3}{3}-\frac{1^3}{3}=\frac{27}{3}-\frac{1}{3}=\frac{26}{3} \end{gathered}[/tex][tex]\begin{gathered} -\int_1^3tdt=-\frac{t^2}{2} \\ evaluate\text{ at 3 and 1} \\ -\frac{3^2}{2}+\frac{1^2}{2}=-\frac{9}{2}+\frac{1}{2}=-4 \end{gathered}[/tex][tex]\begin{gathered} \int_1^36dt=6t \\ evaluate\text{ at 3 and 1} \\ 6(3)-6(1)=12 \end{gathered}[/tex]substitute
[tex]\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt=\frac{26}{3}-4+12=\frac{50}{3}[/tex]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.
Simplify. Assume that all variables result in nonzero denominators.
2n^3 y−8n^2 y/3y^4 * 12/n-4
The simplified form of given expression 2n^3 y−8n^2 y/3y^4 * 12/n-4 is 8n^2/y^3
In this question, we have been given an expression.
2n^3 y−8n^2 y/3y^4 * 12/n-4
We need to simplify given expression.
2n^3 y − 8n^2 y/3y^4 * 12/n-4
= [2n^2y (n - 4)] / 3y^4 * 12/(n - 4)
= 4 * (2n^2y)/y^4
= 8n^2/y^3
Therefore, the simplified form of given expression 2n^3 y−8n^2 y/3y^4 * 12/n-4 is 8n^2/y^3
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If you have a 77.2% and you got 34% on a test and it’s worth 60% of your grade, what would you grade be now?
Answer:
51.28%
Step-by-step explanation:
since the test is worth 60% of your grade, the rest is worth 40%
calculate your new grade by multiplying each grade percent (as written) by the percent of your grade (as a decimal):
77.2(0.4) = 30.88
34(0.6) = 20.4
then add them together: 30.88 + 20.4
Write the expression in the standard form a + bi.
SOLUTION
Write out the expression
[tex]i^{22}[/tex][tex]\begin{gathered} i^{22} \\ \text{can be written as} \\ (i^2)^{11} \end{gathered}[/tex]Recall that
[tex]i^2=-1[/tex]Replace into the expression above
[tex](-1)^{11}=-1[/tex]Hence
[tex]i^{22}=-1[/tex]Therefore
The first option is Right
you are selling snacks at the border trade fair. you are selling nachos and lemonade. each nachos costs $2.50 and each lemonade cost $2.25. at the end of the night you made a total of $112.50. you sold a total of 94 nachos and lemonade combined. how many nachos and lemonades were sold?
In order to determine the number of nachos and lemonade sold, you first write the given situation in an algebraic way.
If x is the number of nachos and y the number of lemonades, then, you have:
2.50x + 2.25y = 112.50 cost of the nachos and lemonade sold
x + y = 94 nachos and lemonade sold
Next, solve the previous system.
Multiply the second equation by 2.50. Next, subtract the equation to the first one:
(x + y = 94)(2.50)
2.50x + 2.50y = 235
2.50x + 2.25y = 112.50
-2.50x - 2.50y = -235
-0.25y = -122.5
solve the previous equation for y:
y = -122.5/(-0.25)
y = 490
Next, replace the previous value of y into the expression x + y = 94 and solve for x:
x + y = 94
x + 490 =
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]which measurement could create more than one triangle measuring 20 cm / 9 cm and 10cm be a triangle with sides measuring 10 cm and 20 cm and included angle measurement 65 C a right angle with acute angles measuring 45 and 45 d a triangle with sides measuring 15 in 20 in and 25 in
Input data
The triangles created by the measurements of options A, B and D have specific side lengths. Therefore, you cannot create more than one triangle.
However, for a triangle with acute angles measuring 45° and 45°, a countless number of similar triangles (triangles with the same shape but different sizes) can be created.
The correct choice is C.
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]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
If 20 assemblers can complete a certain job in 6 hours, how long will the same job take if the number of assemblers is cut back to 8?
ANSWER
[tex]15[/tex]EXPLANATION
For 1 assembler, it will take;
[tex]\begin{gathered} 20\times R\times6=1 \\ R=\frac{1}{120} \end{gathered}[/tex]For 8 assemblers;
[tex]8\times R\times T=1[/tex]Substitute R
[tex]\begin{gathered} 8\times R\times T=1 \\ 8\times\frac{1}{120}\times T=1 \\ \frac{8T}{120}=1 \\ 8T=120 \\ T=\frac{120}{8} \\ =15 \end{gathered}[/tex]Question content area topPart 1A medical researcher administers an experimental medical treatment to patients. The patients in the study are categorized by blood types A, B, AB, and O. The researcher observed that the treatment had a favorable outcome for of the patients with blood type A, of the patients with blood type B, of the patients with blood type AB, and none of the patients with blood type O. Use this information to complete parts (a) through (d).
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
total patients = 300
type A:
total patients = 90
favourable patients = 27
type B:
total patients = 124
favourable patients = 31
type AB:
total patients = 6
favourable patients = 6
type O:
total patients = 80
favourable patients = 0
Step 02:
empirical probability:
probability = favourable outcomes / total outcomes
probability (A) = 27/ 90 = 0.3
probability (B) = 31 / 124 = 0.25
probability (AB) = 6 / 6 = 1
probability (O) = 0 / 80 = 0
That is the full solution.
17. In trapezoid FGJK, what is the value of x? N CO 18.6 L K 23.6 9.3 11.8 ET O 13.6
Given data:
The given figure is shown.
The expression for the trapezium is,
[tex]\begin{gathered} \frac{x}{18.6}=\frac{18.6}{23.6} \\ 23.6x=18.6^2 \\ x=14.6 \\ =15 \end{gathered}[/tex]Thus, thi
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.
Determine the equation of the graphed circle below!Equation should look like the example below!
Step 1:
Write the formula for the equation of a circle.
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ \text{Center = ( a , b )} \\ \text{Radius = r} \end{gathered}[/tex]Step 2:
Locate and write the center and radius of the circle.
Step 3:
Write the equation of the circle with center (-7, -2) and radius r = 2
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-(-7))^2+(y-(-2))^2=2^2 \\ (x+7)^2+(y+2)^2=\text{ 4} \end{gathered}[/tex]Final answer
[tex](x+7)^2+(y+2)^2=\text{ 4}[/tex](X^-3y^2/x^3)^-2
Simplify the expression. Your final answer should use positive exponents.
Answer:
y^-4
here you are
,........
I need help with triangles
Perform the indicated operation -27÷-9
-27/9 = -3
answer is -3
what is the surface area, in square centimeters, of the pyramid ?