The equation of line is [tex]6y=6x$-$12[/tex].
The given slope is [tex]\frac{1}{6}[/tex].
The [tex]y $-$[/tex]intercept is [tex](0, $-$2)[/tex].
We have to write the equation of line using the given slope and [tex]y $-$[/tex]intercept.
The equation of line with the slope m and [tex]y $-$[/tex]intercept of [tex](0,a)[/tex] is [tex]y=mx+a[/tex].
From the question,
The value of [tex]m=\frac{1}{6}[/tex]
The value of [tex]a= $-$2[/tex]
Now putting the value of [tex]m[/tex] and [tex]a[/tex] in the equation of line.
[tex]y=\frac{1}{6}x+( $-$2)\\y=\frac{1}{6}x$-$2[/tex]
Multiply by [tex]6[/tex] on both side
[tex]y\times6=6\times(\frac{1}{6}x$-$2)\\6y=6\times\frac{1}{6}x$-$6\times2\\6y=6x$-$12[/tex]
The equation of line is [tex]6y=6x$-$12[/tex].
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Which points sre vertices of the pre-image, rectangle ABCD?Makes no sense
Given rectangle A'B'C'D', you know that it was obtained after translating rectangle ABCD using this rule:
[tex]T_{-4,3}(x,y)[/tex]That indicates that each point of rectangle ABCD was translating 4 units to the left and 3 units up, in order to obtain rectangle A'B'C'D'.
Notice that the coordinates of the vertices of rectangle A'B'C'D' are:
[tex]\begin{gathered} A^{\prime}(-5,4) \\ B^{\prime}(3,4) \\ C^{\prime}(3,1) \\ D^{\prime}(-5,1) \end{gathered}[/tex]Therefore, in order to find the coordinates of ABCD, you can add 4 units to the x-coordinate of each point and subtract 3 units to each y-coordinate of each point. You get:
[tex]\begin{gathered} A=(-5+4,4-3)=(-1,1) \\ B=^(3+4,4-3)=(7,1) \\ C=(3+4,1-3)=(7,-2) \\ D=(-5+4,1-3)=(-1,-2) \end{gathered}[/tex]Hence, the answers are:
- First option.
- Second option.
- Fourth option.
- Fifth option.
What is the solution to the equation below? 6x= x + 20 O A. x = 4 B. X = 20 C. x = 5 D. No Solutions
Simplify the equation 6x = x +20 to obtain the value of x.
[tex]\begin{gathered} 6x=x+20 \\ 6x-x=20 \\ 5x=20 \\ x=\frac{20}{5} \\ =4 \end{gathered}[/tex]So answer is x = 4
Option A is correct.
**Line m is represented by the equation -2x + 4y = 16. Line m and line k are Blank #1:
Line m:
[tex]y=\frac{2}{3}x+4[/tex]line k:
[tex]\begin{gathered} -2x+4y=16 \\ 4y=2x+16 \\ y=\frac{2x+16}{4} \\ y=\frac{x}{2}+4 \end{gathered}[/tex]so, the lime m and line k are:
D. Neither parallel nor perpendicular
Because:
D. their slopes have no relationship
Find the slope of the graph of the function at the given point.
Consider the following function:
[tex]f(x)=\text{ }\tan(x)\text{ cot\lparen x\rparen}[/tex]First, let's find the derivative of this function. For this, we will apply the product rule for derivatives:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot\frac{d}{dx}\text{ cot\lparen x\rparen + }\frac{d}{dx}\text{ tan\lparen x\rparen }\cdot\text{ cot\lparen x\rparen}[/tex]this is equivalent to:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot(\text{ - csc}^2\text{\lparen x\rparen})\text{+ \lparen sec}^2(x)\text{\rparen}\cdot\text{ cot\lparen x\rparen}[/tex]or
[tex]\frac{df(x)}{dx}=\text{ -}\tan(x)\cdot\text{ csc}^2\text{\lparen x\rparen+ sec}^2(x)\cdot\text{ cot\lparen x\rparen}[/tex]now, this is equivalent to:
[tex]\frac{df(x)}{dx}=\text{ -2 csc \lparen2x\rparen + 2 csc\lparen2x\rparen = 0}[/tex]thus,
[tex]\frac{df(x)}{dx}=0[/tex]Now, to find the slope of the function f(x) at the point (x,y) = (1,1), lug the x-coordinate of the given point into the derivative (this is the slope of the function at the point):
[tex]\frac{df(1)}{dx}=0[/tex]Notice that this slope matches the slope found on the graph of the function f(x), because horizontal lines have a slope 0:
We can conclude that the correct answer is:
Answer:The slope of the graph f(x) at the point (1,1) is
[tex]0[/tex]Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 29 L per minute. There are 400 L in the pond to start. Let W represent the total amount of water in the pond (in liters) and let T represent the total number of minutes that water has been added.Write an equation relating W to T. Then use this equation to find the total amount of water after 13 minutes.Equation : Total amount of water after 13 minutes : liters
In this problem, we have a linear equation of the form
W=mT+b ----> equation in slope-intercept form
where
m is the unit rate or slope of the linear equation
m=29 L/min ----> given
b is the initial value
b=400 L ----> given
substitute
W=29T+400 -------> equation relating W to T.For T=13 min
substitute
W=29(13)+400
W=777 L
the total amount of water after 13 minutes is 777 LIf the radius of both of the green circles is 10 cm, find the area of the yellow region (outside of the circles but inside the rectangle)
The area of the yellow region if the radius of each of the circles is 10 cm is calculated as: 171.7 cm².
How to Find the Area of Circles and Rectangles?The formula that is used to find the areas of circles and rectangles are given below:
Area of a circle = πr², where r is the radius.Area of a rectangle = length × width.Given the diagram in the attachment which shows the green circles and the rectangle, we can deduce the following:
Radius of the each of the circles (r) = 10 cm
Length of the rectangle = 4(r) = 4(10) = 40 cm
Width of the rectangle = 2(r) = 2(10) = 20 cm
The area of the yellow region = area of the rectangle - area of the 2 circles
= (length × width) - 2(πr²)
Substitute
The area of the yellow region = (40 × 20) - 2(π × 10²)
= 800 - 628.3
= 800 - 628.3
= 171.7 cm²
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Describe the relationship between the number 2 x 10^4 and 4 x 10^6
2 * 10^4 = 2 * 10000 = 20,000
4 * 10^6 = 4 * 1000000 = 4,000,000
4,000,000/20,000 = 200
Therefore, 4 * 10^6 is 200 times 2 * 10^4
For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it is not raining. If it is raining, concert organizers estimatethat 7000 people will attend. On the day of the concert, meteorologists predict a 60% chance of rain. Determine the expected number of people who will attend thisconcert
Step 1
Given;
For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it's not raining.
If it is raining, concert organizers estimate 7000 people will attend.
On the day of the concert, meteorologists predict a 60% chance of rain.
Step 2
Given that the probability of having rain is 60%
[tex]Pr(rain)=\frac{60}{100}=0.6[/tex]So the probability of not having rain is;
[tex]\begin{gathered} Pr(rain)+Pr(no\text{ rain\rparen=1} \\ Pr(no\text{ rain\rparen=1-Pr\lparen rain\rparen} \\ Pr(no\text{ rain\rparen=1-0.6=0.4} \end{gathered}[/tex]Step 3
Now, the expected number of people who will attend the concert will be:
=(probability of not having rain x number of expected guests when it does not rain) + (probability of having rain x number of expected guests when rains)
[tex]\begin{gathered} Pr(expected\text{ number of peope\rparen=\lparen0.4}\times11000)+(0.6\times7000) \\ Pr(expected\text{ number of peope\rparen=4400+4200=8600} \end{gathered}[/tex]Answer; So, the expected number of people who will attend the concert is 8600
The odds in favor of a horse winning a race are 7:4. Find the probability that the horse will win the race.A. 7/12B. 4/7C. 7/11D. 4/11
We have a reason for 7:4,
i.e. the total probability of winning is 7+4=11
If the horse has a probability of winning of 7 between 11
We can say that the Pw of the horse is as follows
[tex]\frac{7}{11}[/tex]The answer is the option C
Find the volume of the given solid.Round to the nearest 10th, If necessary. In cubic inches
ANSWER
33.5 cubic inches
EXPLANATION
This is a cone with radius r = 2 in and height h = 8 in. The volume of a cone is,
[tex]V=\frac{1}{3}\cdot\pi\cdot r^2\cdot h[/tex]Replace the known values and solve,
[tex]V=\frac{1}{3}\cdot\pi\cdot2^2in^2\cdot8in=\frac{32}{3}\pi\text{ }in^3\approx33.5\text{ }in^3[/tex]Hence, the volume of the cone is 33.5 in³, rounded to the nearest tenth.
identify point in region of inequalities
We want to picture the inequalities
[tex]y<\text{ - x -3}[/tex]and
[tex]y>\frac{4}{5}x\text{ +5}[/tex]First, we consider the lines y= -x -3 and and y=(4/5) x +5 . Since the first line has a negative slope, this means that its graph should go downwards as x increases and since the other line has a positive slope, this means that its graph should go upwards as x increases. This leads to the following picture
Now, the expression
[tex]y<\text{ -x -3}[/tex]means that the y coordinate of the line should be below the red line. Also, the expression
[tex]y>\frac{4}{5}x+5[/tex]means tha the y coordinate should be above the blue line. If we combine both conditions, we find the following region
so we should look for a point that lies in this region
We are given the points (-1,9), (-6,2), (9,-9) and (-8,-5).
We see that the yellow region is located where the x coordinate is always negative. So, this means that we discard (9,-9).
so we should test the other points. Since -8 is the furthest to the left, let us calculate the value of each line at x=-8.
[tex]\text{ -(-8) -3 = 8 -3 = 5}[/tex]so, in this case the first expression is accomplished since -5 < 5. And
[tex]\frac{4}{5}\cdot(\text{ -8)+5= =}\frac{\text{ -7}}{5}=\text{ -1.4}[/tex]However note that -5 < 1.4, and it should be greater than -1.4 to be in the yellow region. So we discard the point (-8,-5) .
We can check , iusing the graph, that the lines cross at the point (-40/9, 13/9) which is about (-4.44, 1.44). This means that for the point to be on the yellow region, it should be on the left of -4.44. Since the only point that we are given that fulfills this condition is (-6, 2), this should be our answer. We check that
[tex]\text{ -(-6)-3=3>2}[/tex]and
[tex]\frac{4}{5}\cdot(\text{ -6)+5 = }\frac{1}{5}=0.2<2[/tex]so, the point (-6,2) is in the yellow region
Calculate the amount of money that was loaned at 4.00% per annum for 2 years if the simple interest charged was $1,240.00.
Given:-
Simple intrest is $1240. Rate is 4.00%. Time is 2 years.
To find:-
The principal amount.
The formula which relates Simple intrest, Rate, Time and Principal amount is,
[tex]I=prt[/tex]So from this the formula for p is,
[tex]p=\frac{I}{rt}[/tex]Subsituting the known values. we get,
[tex]\begin{gathered} p=\frac{I}{rt} \\ p=\frac{1240}{0.04\times2} \\ p=\frac{1240}{0.08} \\ p=\frac{124000}{8} \end{gathered}[/tex]By simplifying the above equation. we get the value of p,
[tex]\begin{gathered} p=\frac{124000}{8} \\ p=\frac{31000}{2} \\ p=15500 \end{gathered}[/tex]So the principle amount value is 15500.
what is the image of 2,10 after a dilation by a scale factor of 1/2 centered at the origin
A dilation is given by:
[tex](x,y)\rightarrow(kx.ky)[/tex]where k is the scale factor.
In this case we have:
[tex](2,10)\rightarrow(\frac{1}{2}\cdot2,\frac{1}{2}\cdot10)=(1,5)[/tex]Therefore the image is the point:
[tex](1,5)[/tex]3/4 square foot in 1/2 hour what is the unit rate as mixed number
Answer:
[tex]\text{1 }\frac{1}{2}[/tex]Explanation:
The unit rate is:
3/4 divided by 1/2
[tex]\begin{gathered} \frac{3}{4}\times\frac{2}{1} \\ \\ =\frac{3}{2} \end{gathered}[/tex]As a mixed fraction, it is
[tex]\text{1 }\frac{1}{2}[/tex]Because of damage, a computer company had 5 tablets returned out of the 80 that were sold. Suppose the number of damaged tablets sold continue at this rate. How many tablets should the company expect to have returned if it sells 400 of them?
we are told that there 5 damaged tablets out of 80 that are sold. Therefore, the rate of damaged tablets per sold tablets is:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}[/tex]Multiplying this rate by the 400 sold tablets we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}[/tex]Solving we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}=25\text{ damaged}[/tex]Therefore, if the rate continues, the company can expect to return 25 tablets.
Kiran is solving 2x-3/x-1=2/x(x-1) for x, and he uses these steps.He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
Solution:
Given:
[tex]\begin{gathered} To\text{ solve,} \\ \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \end{gathered}[/tex]Kiran multiplied the left-hand side of the equation by (x-1) and multiplied the right-hand side of the equation by x(x-1).
That was where he made the mistake. He ought to have multiplied both sides with the same quantity (Lowest Common Denominator) so as not to change the actual value of the question.
Multiplying both sides by the same quantity does not change the real magnitude of the question.
The actual solution goes thus,
[tex]\begin{gathered} \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \\ \text{Multiplying both sides of the equation by the LCD,} \\ \text{The LCD is x(x-1)} \\ x(x-1)(\frac{2x-3}{x-1})=x(x-1)(\frac{2}{x(x-1)}) \\ x(2x-3)=2 \\ \text{Expanding the bracket,} \\ 2x^2-3x=2 \\ \text{Collecting all the terms to one side to make it a quadratic equation,} \\ 2x^2-3x-2=0 \end{gathered}[/tex]Solving the quadratic equation;
[tex]\begin{gathered} 2x^2-3x-2=0 \\ 2x^2-4x+x-2=0 \\ \text{Factorizing the equation,} \\ 2x(x-2)+1(x-2)=0 \\ (2x+1)(x-2)=0 \\ 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ \text{Dividing both sides by 2,} \\ x=-\frac{1}{2} \\ \\ \\ OR \\ x-2=0 \\ x=0+2 \\ x=2 \end{gathered}[/tex]Therefore, the actual solutions to the expression are;
[tex]\begin{gathered} x=-\frac{1}{2} \\ \\ OR \\ \\ x=2 \end{gathered}[/tex]Can I have help with this problem? I don't really understand how to graph this
Step 1:
The graph of y = -2 is a horizontal line passing through -2.
Step 2
The table shows the fraction of students from differentgrade levels who are in favor of adding new items tothe lunch menu at their school. Which list shows the grade levels in order from the greatest fraction of students to the least fraction of students ?
First, write all the fractions using the same denominator. To do so, find the least common multiple of all denominatos. The denominators are:
[tex]50,20,25,75,5[/tex]The least common multiple of all those numbers is 300.
Use 300 as a common denominator for all fractions to be able to compare their values.
5th grade
[tex]\frac{33}{50}=\frac{33\times6}{50\times6}=\frac{198}{300}[/tex]6th grade
[tex]\frac{13}{20}=\frac{13\times15}{20\times15}=\frac{195}{300}[/tex]7th grade
[tex]\frac{18}{25}=\frac{18\times12}{25\times12}=\frac{216}{300}[/tex]8th grade
[tex]\frac{51}{75}=\frac{51\times4}{75\times4}=\frac{204}{300}[/tex]9th grade
[tex]\frac{3}{5}=\frac{3\times60}{5\times60}=\frac{180}{300}[/tex]Now, we can compare the numerators to list the fraction from greatest to lowest:
[tex]\begin{gathered} \frac{216}{300}>\frac{204}{300}>\frac{198}{300}>\frac{195}{300}>\frac{180}{300} \\ \Leftrightarrow\frac{18}{25}>\frac{51}{75}>\frac{33}{50}>\frac{13}{20}>\frac{3}{5} \\ \Leftrightarrow7th\text{ grade}>8th\text{ grade}>5th\text{ grade}>6th\text{ grade}>9th\text{ grade} \end{gathered}[/tex]Therefore, the list of grade levels in order from the greatest fraction of students to the least fraction of students, is:
7th grade (18/25)
8th grade (51/75)
5th grade (33/50)
6th grade (13/20)
9th grade (3/5)
A box of a granola contains 16.8 ounces . It cost $5.19 . What is the cost , to the nearest cent , of the granola per ounce ? A . $0.12 B . $0.31 C . $3.24
The cost per unit ounce is obtained by computing the quotient:
[tex]c=\frac{C}{N}.[/tex]Where:
• c is the cost per unit ounce,
,• C is the cost,
,• N is the number of ounces that you get for C.
In this problem we have:
• C = $5.19,
,• N = 16.8 ounces.
Computing the quotient, we get:
[tex]c=\frac{5.19}{16.8}\cong0.31[/tex]dollars per ounce.
Answer: B. $0.31
A survey of 100 high school students provided thisfrequency table on how students get to school:Drive toTake theGradeWalkSchoolbusSophomore2253Junior13202Senior2555Find the probability that a randomly selected studenteither takes the bus or walks.[?P(Take the bus U Walk)
Let's call the event of a student taking the bus as event A, and the event of a student walking as event B. The theoretical probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes. We have a total of 100 students, where 50 of them take the bus and 10 of them walk. This gives to us the following informations:
[tex]\begin{gathered} P(A)=\frac{50}{100} \\ P(B)=\frac{10}{100} \end{gathered}[/tex]The additive property of probability tells us that:
[tex]P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B)[/tex]Since our events are mutually exclusive(the student either walks or takes the bus), we have:
[tex]P(A\:and\:B)=0[/tex]Then, our probability is:
[tex]P(A\cup B)=\frac{50}{100}+\frac{10}{100}-0=\frac{60}{100}=\frac{3}{5}[/tex]The answer is:
[tex]P(Take\:the\:bus\cup Walk)=\frac{3}{5}[/tex]I need help with this practice problem Having a tough time solving properly
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
r = 7 sin (2θ)
Step 02:
polar equation:
r = 7 sin (2θ):
r = a sin nθ
n odd ==> n petals
n even ===> 2n petals
n = 2 ===> 2*2 petals = 4 petals
graph:
length of the petals:
r = 7 sin (2θ)
θ = 45°
r = 7 sin (2*45°) = 4.95
The answer is:
4.95
are figures A and B congruent? explain your reason
if x=10 units, then what is the volume of the cube
Knowing that the solid is a cube, you can use the following formula for calculate its volume:
[tex]V=s^3[/tex]Where "s" is the length of any edge of the cube.
In this case, you can identify that:
[tex]s=x=10units[/tex]A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the building? explain your reasoning.show your work. round to the nearest tenth if necessary.
The ladder must be 9.4 ft to reach the top of the building
Here, we want to get the length of the ladder that will reach the top of the building
Firstly, we need a diagrammatic representation
We have this as;
As we can see, we have a right triangle with the hypotenuse being the length of the ladder
We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides
Thus, we have;
[tex]\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}[/tex]Larry purchased a new combine that cost $260,500, minus a rebate of $5,500, a trade-in of $8,500, and a down payment of $7,000. He takes out a loan for the balance at 8% APR over 4 years. Find the annual payment. (Simplify your answer completely. Round your answer to the nearest cent.)
The annual payment for the loan balance is $72,310.03.
What is the periodic payment?The periodic payment is the amount that is paid per period (yearly, monthly, quarterly, or weekly) to repay a loan or a debt.
The periodic payment can be computed using an online finance calculator, making the following inputs.
N (# of periods) = 4 years
I/Y (Interest per year) = 8%
PV (Present Value) = $239,500 ($260,500 - $5,500 - $8,500 - $7,000)
FV (Future Value) = $0
Results:
PMT = $72,310.03
Sum of all periodic payments = $289,240.13
Total Interest = $49,740.13
Thus, the annual payment that Larry needs to make is $72,310.03.
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Question #3 3) The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get 99. Find the original number.
ones number = x
Tens number = y
y>x
Number at the tens place y = (x+5)
original number = 10 (x+5)+x
Interchange digits:= 10x+(x+5)
original number + new number = 99
¨[10(x+5)+x]+ [10x+ (x+5)] =99
Solving for x:
(10x+50+x )+( 10x+x+5) = 99
Combine like terms
(11x+50) + (11x+5) = 99
11x+11x+50+5 =99
22x+55 =99
subtract 55 from both sides
22x +55-55= 99-55
22x = 44
Divide both sides by 22
22x/22= 44/22
x = 2
unit place: 2
tens place = x+5 = 2+5 = 7
original number = 72
P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational function [y=P(x)/q(x)], draw a graph and answer the following: a) what are the zeroes?b) are there any asymptotes? c) what is the domain and range for this function?d) it it a continuous function?e) are there any values of y= f(x)/g(x) that are undefined? Explain
we have the following function
[tex]\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}[/tex]where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.
We will first draw the function
from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis
In this case, we see that that the only zero is at x=0.
Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.
Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.
From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous
Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined
si f(x) = x + 5 cuanto es f(2) f(1) f(0) f(-1) f-(-2) f(a)
f (x)= x+ 5
f(2)
Reemplaza x por 2 y resuelve
f(2)= 2 + 5 = 7
Mismo procedimiento para los demas valores:
f(1) = 1 + 5 = 6
f(0) = 0 + 5 = 5
f(-1)= -1+5 = 4
f(-2)= -2+5 = 3
f(a)= a + 5
For the diagram below, if < 4 = 4x - 2, and < 6 = 2x + 14, what is the value of x?Select one:a.8b.16c.4d.5
x = 8
ExplanationsFrom the line geometry shown, the line a and b are parallel lines while line "t" is the transversal.
Since the horizontal lines are parallel, hence;
[tex]\angle4=\angle6(alternate\text{ exterior angle})[/tex]Given the following parameters
[tex]\begin{gathered} \angle4=4x-2 \\ \angle6=2x+14 \end{gathered}[/tex]Equate both expressions to have:
[tex]\begin{gathered} 4x-2=2x+14 \\ 4x-2x=14+2 \\ 2x=16 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Hence the value of x is 8
A 5p coin weighs 4.2g. Approximately, how much will one million pounds worth of 5p pieces
weigh?
Answer:
It would weight 840,000g
Step-by-step explanation:
1,000,000 ÷ 5
= 200,000
= 200,000 × 4.2
= 840,000