Which set of expressions are not equivalent?(4 1\2 a + 2) + 3 and 5 + 4 a18 a + 14 and 6 a + 12 + 2 + 1 2a0.5 a + 3.5 + 4 a and 3.5 + 4.5 a3 a + 7 +2 a and 7 a + 5
Solution:
Solution:
To find which of given expressions are equivalent
For the first expressions
[tex](4\frac{1}{2}a+2)+3\text{ and 5}+4\frac{1}{2}a[/tex]Solving the expression
[tex]\begin{gathered} 4\frac{1}{2}a+2+3=5+4\frac{1}{2}a \\ 5+4\frac{1}{2}a=5+4\frac{1}{2}a \end{gathered}[/tex]The expressions are equivalent
For the second expression
[tex]18a+14\text{ and 6a}+12+2+12a[/tex]Solving the expression
[tex]\begin{gathered} 18a+14=6a+12a+12+2 \\ 18a+14=18a+14 \end{gathered}[/tex]The expressions are equivalent
For the third expression
[tex]0.5a+3.5+4a\text{ and 3.5}+4.5a[/tex]Solving the expression
[tex]\begin{gathered} 3.5+0.5a+4a=3.5+4.5a \\ 3.5+4.5a=3.5+4.5a \end{gathered}[/tex]The expressions are equivalent
For the fourth expression
[tex]3a+7+2a\text{ and 7a}+5[/tex]Solving the expression
[tex]\begin{gathered} 3a+2a+7=7a+5 \\ 5a+7\ne7a+5 \end{gathered}[/tex]The expressions are not equivalent.
Hence, the set of expressions that are not equivalent is
[tex]3a+7+2a\text{ and 7a}+5[/tex]Helena is watching a movie on television she notices that there are 9 1/4 min of commercial for every 1/2 hour of the movie
If Helena watches a movie that shows a commercial for 9 1/4 mins of every 30 minutes (half hour), then the unit rate of commercials per hour is derived as;
[tex]\begin{gathered} \text{Com /30 mins=9}\frac{1}{4} \\ \text{Com /60 mins=9}\frac{1}{4}\times2 \\ \text{Com /hour=}\frac{37}{4}\times2 \\ \text{Com /hour=}\frac{37}{2} \\ Com\text{ /hour=18}\frac{1}{2} \end{gathered}[/tex]The unit rate of commercials per hour is 18 1/2 minutes (eighteen and half minutes)
The measures of 6 interior angles of a heptagon are: 111, 110, 121, 135, 139 and 92.Find the measure of the seventh interior angle. show work please:)
The sum of the interior angles of a heptagon 900º, if you know 6 of the 7 interior angles of the heptagon, you can determine the measure of the seventh angle as follows:
[tex]\begin{gathered} 900=111+110+121+135+139+92+x \\ 900=708+x \end{gathered}[/tex]Subtract 708 from both sides of the equal sign to reach the measure of the seventh angle:
[tex]\begin{gathered} 900-708=708-708+x \\ 192=x \end{gathered}[/tex]The seventh interior angle has a measure of 192º
Susan works as a Hostess where she makes $8.75 per hour she also babysitting earns $10 per hour last week she made $165 if she works 6 hours more as a Hostess than she did babysitting how many hours did she babysit last week?
6 hours as babysitter last week
1) Gathering the data
Hostess
8.75h
Babysitting
10h
2) Based on that, we can write out the following equation, considering that Susan worked 6 hours more:
8.75(h+6) +10h= 165 Distribute the factor 8.75
8.75h +52.5 +10h = 165 Combine like terms
18.75h = 165 -52.5 Subtract from both sides 52.5
18.75h = 112.5 Divide both sides by 18.75
h =6
3) Hence, we can conclude that Susan worked as a hostess for 12 hours and as a baby sitter 6 hours last week
what sentence represent the number of poins in the problem below
hello
to solve this question, we can write out two sets of equation and solve them
let the short answers be represented by x
let the multiple-choice questions be represented by y
we know that the test has 60 points
multiple-choice carries 2 points
short answers carries 5 points
[tex]2x+5y=60[/tex]now we have a total of 15 questions which comprises of multiple-choice questions and short answers
[tex]x+y=15[/tex]now we have two set of equations which are
[tex]\begin{gathered} 2x+5y=60\ldots\text{equ}1 \\ x+y=15\ldots\text{equ}2 \end{gathered}[/tex]now let's solve for x and y
from equation 2, let's make x the subject of formula
[tex]\begin{gathered} x+y=15 \\ x=15-y\ldots\text{equ}3 \end{gathered}[/tex]put equation 3 into equation 1
[tex]\begin{gathered} 2x+5y=60 \\ x=15-y \\ 2(15-y)+5y=60 \\ 30-2y+5y=60 \\ 30+3y=60 \\ \text{collect like terms} \\ 3y=60-30 \\ 3y=30 \\ \text{divide both sides by the coeffiecient of y} \\ \frac{3y}{3}=\frac{30}{3} \\ y=10 \end{gathered}[/tex]now we know the value of y which is the number of multiple-choice question. we can use this information to find the number of short answer through either equation 1 or 2
from equation 2
[tex]\begin{gathered} x+y=15 \\ y=10 \\ x+10=15 \\ \text{collect like terms} \\ x=15-10 \\ x=5 \end{gathered}[/tex]from the calculations above, the number of short answers is equal to 5 and multiple-choice questions is equal to 10.
The answer to this question is option C
if you can't read it, it says : Add -7d-3 and 10d-6. Show all steps
-7 d - 3 + 10d - 6
First step
Group in parenthesis and
Add both letter terms
(-7d + 10d)= -3d
Second step
Group and add numbers only
(-3 - 6) = -9
Now add both results
=-3d - 9
Answer is -3d - 9
the statement below always sometimes or never true give at least two examples to support your reasoning the LMC of two numbers is the product of the two numbers discuss with a partner
The distance to the grocery store is 13.456 miles round this distance to the nearest tenth
Answer:
Your answer is 13.50
Mimstoon started with at most 2 boondins (y). Every day (x), he
bought at most 1/2 more of them.
Write an inequality to model this relationship.
Step-by-step explanation:
y <= x/2 + 2
the maximum is to have 2 boondins and add 1/2 boondin every day.
but every time Mimstoon did not use the max. possible, the resulting total sum of boodins stays smaller that the maximum, and is therefore a valid data point.
(6, 7) is not in the inequality relationship.
because the max. boondins after 6 days is 2 (from the beginning) and 1/2 every day = 2 + 1/2 × 6 = 2 + 3 = 5.
but the data point shows 7 as y value (which is larger than the allowed max. of 5).
therefore the inequality is false for this data point, and therefore the data point is not in the inequality relationship.
which trig function should you pick ? solve for x
To find the value of x and according to the given information, it is necessary to use cosine, which is the ratio between the adjacent side and the hypotenuse.
In this case the given angle is 43, it means that cos 43 is the ratio between 36 (adjacent side) and x (hypotenuse). Solve for x, this way:
[tex]\begin{gathered} \sin 43=\frac{36}{x} \\ x=\frac{36}{\sin 43} \\ x=52.8 \end{gathered}[/tex]x has a value of 52.8 (approximately).
Stores on the verbal Graduate Record Exam (GRE) have a mean of 462 and a standard deviation of 119. Scores on the quantitative GRE have a mean of 584 and a standard deviation of 151. Assuming the scores are normally distributed, what quantitative scores are required for an applicant to score at or above the 90th percentile
We need Z-score here.
The formula is:
[tex]z=\frac{x-\mu}{\sigma}[/tex]A normal curve (with 90th percentile), looks like the one below:
We need a standard normal table to move further.
When we go to the table, we find that the value 0.90 is not there exactly, however, the values 0.8997 and 0.9015 are there and correspond to Z values of 1.28 and 1.29, respectively
(i.e., 89.97% of the area under the standard normal curve is below 1.28).
The exact Z value holding 90% of the values below it is 1.282.
--------------- Now, we work backwords and find the value of x:
Verbal GRE:
[tex]\begin{gathered} \mu=462 \\ \sigma=119 \\ z=\frac{x-\mu}{\sigma} \\ 1.282=\frac{x-462}{119} \\ x-462=1.282(119) \\ x=614.558 \end{gathered}[/tex]So, a 90th percentile on Verbal GRE is a score above 614.56
Quantitative GRE:
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ 1.282=\frac{x-584}{151} \\ x=777.582 \end{gathered}[/tex]So, a 90th percentile on Quantitative GRE is a scoer above 777.58
The Jacksons bought a $276,000 condominium. They made a down payment of S43,000 and took out a mortgage for the rest. Over the course of 30 years theymade monthly payments of $1396.96 on their mortgage until it was paid off.Х$?(a) What was the total amount they ended up paying for the condominium(including the down payment and monthly payments)?si(b) How much interest did they pay on the mortgage?si
(a) What was the total amount they ended up paying for the condominium
(including the down payment and monthly payments)?
We need to calculate the total monthly payments, plus the down payment.
[tex]=1396.96\cdot30\cdot12+43000=545905.60[/tex]So, the total amount they ended up paying was $545,905.60
(b) How much interest did they pay on the mortgage?
The interest they pay is: $269,905.60
[tex]=545905.60-276000.00=269905.60[/tex]of the arithmetic sequence -10, -25, -40, ...
Find the 81st term
The 81st term of the arithmetic sequence -10, -25, -40, ... is -1210
Here is a step-by-step explanation The common difference in this sequence,
d = a2 -a1
let a2 = -25 and a1 = -10
d= -25 - (-10)
d = -15
the common difference (d) = -15
To find the n term we use :
an= a1+ (n-1)d
Therefore,
a81 = -10 + (81-1)(-15)
= -1210
The 81stterm of the arithmetic sequence -10, -25, -40, ... -1210
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After filling the ketchup dispenser at the snack bar where she works, Kelley measures the level of ketchup during the day at different hourly intervals.
Complete parts a to c.
Question content area bottom
Part 1
a. Assuming the ketchup is used at a constant rate, write a linear equation that can be used to determine the level of ketchup in the dispenser after x hours. Let y represent the level of ketchup in inches.
y equals negative five eighths x plus 15
(Type an equation.)
Part 2
b. How can the equation from part a be used to find the level of ketchup when the dispenser is full?
A.
Identify the slope of the line represented by the equation.
B.
Find the difference between the slope and the y-intercept of the line represented by the equation.
C.
Substitute 0 for x in the equation and solve for y. Equivalently, find the y-intercept.
D.
Substitute 0 for y in the equation and solve for x. Equivalently, find the x-intercept.
Part 3
c. If Kelley fills the ketchup dispenser just before the snack bar opens and the snack bar is open for 18 hours, will the dispenser need to be refilled before closing time? Explain.
If the ketchup is used at a constant rate, then the equation from part a indicates the dispenser will become empty
enter your response here hours after the snack bar opens. This means that the dispenser
▼
will become empty before closing time,
will become empty exactly at closing time,
still has ketchup in it at closing time,
and so it
▼
will not
will
need to be refilled before closing time.
Answer:
Answers
Step-by-step explanation:
Got it from SAVVAS
Find two points on the line.4y + 2x = 7
Given the equation of a line:
4y +2x = 7
It's required to find two points on the line.
Solving for y:
[tex]y=\frac{7-2x}{4}[/tex]We give x any value and calculate the corresponding value of y.
For example, for x = 1:
[tex]y=\frac{7-2(1)}{4}=\frac{5}{4}[/tex]The point is (1, 5/4)
Now for x = -4:
[tex]y=\frac{7-2(-4)}{4}=\frac{15}{4}[/tex]The point is (-4, 15/4)
Which relation is a function? 4 15 O -6- 7 C fy 0 -H O O O -2 -4
The most appropriate choice for functions will be given by -
Third option is correct
Third relation is a function.
What is a function?
A function from A to B is a rule that assigns to each element of A a unique element of B. A is called the domain of the function and B is called the codomain of the function.
There are different operations on functions like addition, subtraction, multiplication, division and composition of functions.
Here,
For a function a point in the domain has a unique image.
Here values of x axis represents domain and values of y axis represents the range
For the first option,
x = -1 has two images, y = -1 and y = 3
so x = -1 do not have a unique image
So the first relation is not a function
For the second option,
x = 0 has two images, y = -1 and y = 2
so x = 0 do not have a unique image
So the second relation is not a function
For the third option,
Every point of the domain has a unique image
So the third relation is a function
For the fourth option,
x = -2 has two images, y = 1 and y = -2
so x = -2 do not have a unique image
So the fourth relation is not a function
so, Third option is correct
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Solve the quadratic equation by completing the square.x^2-14x+46=0First choose the appropriate form and fill in the blanks with the correct numbers. Then solve the equation. If there’s more than one solution, separate them with commas.
x² - 14x + 46 = 0
subtract 46 from both-side
x² - 14x = -46
Add the square of the half of the co-efficient of x
x² - 14x + (-7)² = -46 + 7²
(x-7)² = -46 + 49
(x-7)² =3
Take the square root of both-side
x-7 = ±√3
x-7= ±1.732
Add 7 to both-side of the equation.
x= 7 ± 1.732
Eithe x= 7 + 1.732 or x= 7 - 1.732
x=8.732 or x=5.268
Therefore x = 8.732 , 5.268
A 17 m guy wire attached to the top of a tower (the height of the tower is not yet known) is anchored on the ground, 8 m away from the base of the tower. A second guy wire needs to be attached to the centre of the tower and then anchored to the same ground-anchor as the first wire.Draw and label a diagram.How long does the second guy wire need to be?Determine the measure of the angle formed between the two wires.
Solution
Step 1:
Draw the diagram to illustrate the information.
Step 2:
Use the Pythagoras theorem to find the height of the pole.
[tex]\begin{gathered} 17^2\text{ = h}^2\text{ + 8}^2 \\ 289\text{ = h}^2\text{ + 64} \\ h^2\text{ = 289 - 64} \\ h^2\text{ = 225} \\ h\text{ = }\sqrt{225} \\ \text{h = 15m} \end{gathered}[/tex]Step 3
b) The height of the second guy wire = d
[tex]\begin{gathered} \text{Apply the pythagoras theorem} \\ d^2=\text{ \lparen}\frac{15}{2}\text{\rparen}^2+\text{ 8}^2 \\ d^2\text{ = 56.25 + 64} \\ d^2\text{ = 120.25} \\ \text{d = }\sqrt{120.25} \\ \text{d = 10.97m} \end{gathered}[/tex]c)
[tex]\begin{gathered} sin(\theta\text{ + }\alpha)\text{ = }\frac{Opposite}{Hypotenuse} \\ sin(\theta\text{ + }\alpha)\text{ = }\frac{15}{17} \\ \theta\text{ + }\alpha\text{ = sin}^{-1}(\frac{15}{17}) \\ \theta\text{ + }\alpha\text{ = 61.9}^o \end{gathered}[/tex][tex]\begin{gathered} sin\alpha\text{ = }\frac{7.5}{10.97} \\ \alpha=\text{ sin}^{-1}(\frac{7.5}{10.97}) \\ \alpha\text{ = 43.1} \end{gathered}[/tex]The angle between the two guys' wires = 61.9 - 43.1
Measure of the angle formed between the two wires = 18.8
Can you help me solve the problem by multiplying or dividing.
Given:
The width (w) of a rectangular swimming pool is
[tex]w=7x^2[/tex]The area (A) of the pool is
[tex]A=7x^3-42x^2[/tex]Required:
The expression for the length (l) of the pool.
Answer:
Let us find the expression for the length (l) of the pool.
The area (A) of the rectangular swimming pool is,
[tex]\begin{gathered} A=l\times w \\ 7x^3-42x^2=l\times7x^2 \\ l=\frac{7x^3-42x^2}{7x^2} \\ l=\frac{7x^3}{7x^2}-\frac{42x^2}{7x^2} \\ l=x-6 \end{gathered}[/tex]Final Answer:
The expression for the length (l) of the pool is,
[tex]l=x-6[/tex]Solve |3x +18| = 6x
Please help
Answer: x = 2, 6
Step-by-step explanation:
To solve, we will isolate the variable. Since this deals with absolute value, we can assume we'll have two answers.
|3x + 18| = 6x
3x + 18 = 6x 3x + 18 = - 6x
18 = 3x 18 = 9x
6 = x 2 = x
x = 2, 6
how to write the standard equation for the hyperbola that is on the graph
The standard form of the hyperbola:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}\text{ = }1[/tex]The center of the hyperbola is at (h, k) = The point where the lines intersect
(h, k) = (0, 0)
For hyperbola:
c² = a² + b²
b is gotten by tracing the value of a to the intersecting line. Trace 3 down to the x axis. you get 5.
[tex]\begin{gathered} a\text{ = distance from center to vertex} \\ \text{the hyperbola coordinate = (0, a) = }(0,\text{ 3)} \\ \text{the other one = (0, -a) = (0, -3)} \\ \text{Hence, a = 3} \end{gathered}[/tex][tex]\begin{gathered} c\text{ = distance from the center to the focus point} \\ b\text{ = 5 and -b = 5} \\ c^2=3^2+5^2 \end{gathered}[/tex][tex]\begin{gathered} c^2\text{ = 9+25 = 34} \\ c\text{ = }\sqrt[]{34} \\ c\text{ = 5.83} \end{gathered}[/tex]The equation becomes:
[tex]\begin{gathered} \frac{(y-k)^2}{5^2}-\frac{(x-h)^2}{3^2}=\text{ }\frac{(y-0)^2}{5^2}-\frac{(x-0)^2}{3^2} \\ =\text{ }\frac{y^2}{25}-\frac{x^2}{9} \end{gathered}[/tex]the cost price of 24 articles is the selling price of 15 articles find the gain percentage
Answer:
Step-By step explanation:
A professional football prospect runs a 40 yard dash in 5 seconds.What is the player's average speed over this distance?a.) 8 yards per secondb.) 0.625 yards per secondc.) 0.125 yards per secondd.) 5 yards per second
. Find the Value of x. SHOW YOUR WORK OR NO CREDITIMI
we can use pythagoras to solve the triangle
[tex]a^2+b^2=h^2[/tex]where a and b are sides and h the hypotenuse on this case we use half triangle to make a rigth triangle and apply pythagoras
a is 9, b is 20 and h the x
so replacing
[tex]\begin{gathered} 9^2+20^2=x^2 \\ \end{gathered}[/tex]solve for x
[tex]x=\sqrt[]{9^2+20^2}[/tex]and do the operatios
[tex]\begin{gathered} x=\sqrt[]{81+400} \\ x=\sqrt[]{481}\approx21.93 \end{gathered}[/tex]the value of x is 21.93 units
how do I solve Tan-1(0.577)?
In the given fexpressionwe have,
Tan-1(0.577)
[tex]\begin{gathered} \text{Tan}^{-1}(0.577) \\ \text{Let x = Tan}^{-1}(0.577) \\ \text{Multiply both side by Tan} \\ \text{Tan x=0.577} \\ \text{ From the trignometric table find at what degr}e\text{ the tangent will be 0.577} \\ \text{ SO, }from\text{ Trignometric table we have, Tan 29.98}=0.577 \\ So,Tan^{-1}(0.577)\text{ = 29.98} \end{gathered}[/tex]Answer : 29.98 degress
Show where the expression in number 8 would be on a unit circle
Solution
Step 1:
Write the expression
[tex]cos270\text{ }[/tex]Step 2
[tex]\begin{gathered} cos270\text{ = 0} \\ sin270\text{ = -1} \\ (0,\text{ -1\rparen} \end{gathered}[/tex]Question 8 Second part
[tex]\begin{gathered} sin(\frac{16\pi}{3})\text{ = sin960 = sin\lparen960 - 2}\times360)\text{ = sin240} \\ sin(\frac{16\pi}{3})\text{ = }\frac{-\sqrt{3}}{2} \\ cos(\frac{16\pi}{3})\text{ = }\frac{-1}{2} \end{gathered}[/tex][tex](\frac{-1}{2},\text{ }\frac{-\sqrt{3}}{2})[/tex]The math class is selling raffle tickets to raise money for the fall ball the raffle tickets cost them $0.08 each at the office supply store and they spent $55 making T-shirts for the club members to wear. they are selling the raffle tickets for two dollars each if X represents the number of raffle tickets they sell ,the following functions can be used to present the situation where c(x) is the clubs cost and r(x) is the amount of money they make selling the tickets
We have the next information
x is the number of raffle tickets
the Profit function is R(x) function minus C(x)
C(x)=0.08x+55
R(x)=2x
P(x)=R(x)-C(x)=2x-(0.08x+55)=2x-0.08x-55
We sum similar terms
P(x)=1.92x-55
I need help with this problem. The table below shows that the number of miles driven by Samuel is directly proportional to the number of gallons he used. \text{Gallons Used}Gallons Used \text{Miles Driven}Miles Driven 3030 10411041 4646 1596.21596.2 4848 1665.61665.6 \text{What is the rate of gas usage, in miles per gallon?} What is the rate of gas usage, in miles per gallon?
The rate of gas usage, in miles per gallon is 34.7 miles per gallon.
What is a direct proportion?A direct proportion is used to show the constant of proportionality.
In this case, at 30 gallons, the miles used was 1041. Therefore the rate of gas usage will be:
= Miles travelled / Number of gallons
= 1041 / 30
= 34.7 miles per gallon
This shows the rate.
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Determine the sum of the measure of the interior angle of the specified polygon (octagon)A/720B/900C/1080D/1440
Solution:
The sum of interior angles of a polygon is calculated using the formula below
[tex]=(n-2)\times180[/tex]Since the polygon is an octagon,
[tex]n=8[/tex]By substituting the values, we will have
[tex]\begin{gathered} (n-2)\times180 \\ (8-2)\times180 \\ =6\times180 \\ =1080^0 \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow1080^0[/tex]OPTION C is the right answer
Suppose you choose four booksto read from a summer readinglist of 12 books. How manydifferent combinations of booksare possible?Note: nCrn!r!(n-r)!
Answer
495 different combinations of books are possible.
Explanation
The different combinations of books possible can be known using the given formula
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex]Choosing four books to read from a list of 12 books means r = 4 and n = 12
Therefore
[tex]\begin{gathered} _nC_r=\frac{n!}{r!(n-r)!}=\frac{12!}{4!(12-4)!}=\frac{12!}{4!\times8!}=\frac{12\times11\times10\times9\times8!}{4\times3\times2\times8!}=495 \\ \\ \end{gathered}[/tex]Hence, there are 495 different combinations of books are possible
