Given data
*The given fraction is 8/10
[tex]\frac{80}{100}=\frac{8}{10}[/tex]80/100 is the fraction equivalent to 8/10
the question is on the sheet .The shaded region is the area within the figure but not the corner squares
Based on the information given in the exercise, you know that the shaded region has this area:
[tex]6x^2-2xy+3y^2[/tex]And the area of each square region in the corners is:
[tex]4x^2[/tex]Since there are four square regions, you can multiply that the area of any of them by 4, in order to find the total area of them:
[tex]4(4x^2)=16x^2[/tex]Then, in order to find the total area of the figure, you must add the area of the shaded region and the total area of the square regions:
[tex](6x^2-2xy+3y^2)+(16x^2)=22x^2^{}-2xy+3y^2[/tex]Therefore, the total of the figure is:
[tex]22x^2-2xy+3y^2[/tex]Fart A Now that you have converted a terminating decimal number Into a fractlon, try converting a repeating decimal number Into a fraction. Repeating decimal numbers are more difficult to convert Into fractions. The first step is to assign the given decimal number to be equal to a varlable, x. For the decimal number 0.3, that means X = 0.3. if x = 0.3, what does 10x equal? Font Sizes
Given x = 0.3, we're asked to find 10x. All we need to do is multiply 10 by 0.3(which is the value of x);
[tex]10\text{ }\ast\text{ 0.3 = 3}[/tex]Therefore, 10x is equal to 3.
Which ordered pair is a solution set for the linear equation, x + 16 = 6y? A (16,4) B (2, 3) C (-4,-2) D (-10,-1)
Explanation
to figure out if a pair is a solution, just replace x and y values and check if the equation is true,then
Step 1
[tex]\begin{gathered} x+16=6y \\ A)(16,4)\rightarrow when\text{ x=16 and y =4} \\ \text{replace} \\ 16+16=6\cdot4 \\ 32=24\rightarrow False,then\text{ A is not a solution} \end{gathered}[/tex]Step 2
[tex]\begin{gathered} B)(2,3)\rightarrow whenx=\text{ 2 and y=3} \\ x+16=6y \\ \text{replace} \\ 2+16=6\cdot3 \\ 18=18\rightarrow true \\ so\text{ B is a solution} \end{gathered}[/tex]Step 3
[tex]undefined[/tex]Rigid Transformations: Question 3A military compound is located in a city where the streetsare arranged in a grid. The compound is going to be movedto a new location in the same city. The new compound willbe exactly the same as the old one and in the sameorientation. The existing location is represented on thecoordinate grid shown, where each unit represents one cityblock. Which of the following best describes the situation?14B1210CDD'C6 4AB2-14 12 10 86226 8 10 12 142D4С6 8N10Wm12s141
In this question, we are given two locations of a military compound.
Let's take the coordinates of point A from the graph,
A = (-3, 2)
The new coordinates of A' = (5, 13)
translation in x-axis = 5 - (-3) = 5 + 3 = 8
translation in y-axis = 13 - 2 = 11
hence, each point of the figure will be translated to 8 units on the x-axis and 11 units on the y-axis.
D(-9,4) E(-3,4) F(-3,10) G(-9,10) rotation 180 clockwise
Answer:
D = (9,-4) E = (3,-4) F= (3, -10) G=(9,-10)
Step-by-step explanation:
Simply switch the signs (- or +)
Ex: rotate (9,1) 180 degrees
Your answer would be (-9,-1)
Jason predicted that 227 students would attend the school dance.
The actual number was 250. What is the percent error of Jason's prediction?
.
1. What is the difference between the predicted value and the
actual value?
2. Complete the equation:__=p • ___
3. Solve the equation for p.
(I already did number one I just need help with 2 and 3)
Question 2
[tex]23=p \cdot 250[/tex]
The difference is 23.The actual value is 250.Question 3
[tex]p=\frac{23}{250}=0.092[/tex]
13. A 640 kg of a radioactive substance decays to 544 kg in 13 hours. A. Find the half-life of the substance. Be sure to show your work including the formulas you used. Round to the nearest tenth of an hour. Only solutions using formulas from the 4.6 lecture notes will receive credit.B. How much of the substance is present after 3 days? Be sure to show the model you used.C. How long does it take the substance to reach 185 kg? Be sure to show your work.
EXPLANATION
The equation for half-life is given by the following formula:
[tex]H=\frac{t\cdot\ln(2)}{\ln(\frac{A_0}{A_t})}[/tex]Replacing terms:
[tex]H=\frac{t\cdot\ln(2)}{\ln(\frac{A_0}{A_t})}=\frac{13\cdot\ln(2)}{\ln(\frac{640}{544})}=\frac{9.0109}{0.1625}=55.45[/tex]The half-life time is H =55.4 hours.
B) After three days, that is, 72 hours, the amount of substance will be given by the following relationship:
[tex]A=A_o\cdot e^{-(\frac{\ln2}{H})t}=640\cdot e^{-(\frac{\ln2}{55.4})\cdot72}=640\cdot e^{-0.90084}[/tex]Multiplying terms:
[tex]A=640\cdot0.4062=259.96\text{ Kg}[/tex]There will be 259.96 Kg after 3 days.
C) In order to compute the number of days that will take to the substance to reach a concentration equal to 185 Kg, we need to apply the following formula:
[tex]t=\frac{\ln (\frac{A}{A_o})}{-\frac{\ln (2)}{t\frac{1}{2}}}[/tex]Replacing terms:
[tex]t=\frac{\ln (\frac{185}{544})}{-\frac{\ln (2)}{55.45}}=\frac{-1.0785}{-0.0125}=\frac{1.0785}{0.0125}=86.28\text{ hours}[/tex]It will take 86.28 hours to the substance to reach 185 Kg.
6. If you start with 200 MNM's and eat 15 every minute and your friend starts with300 MnM's but eats 25 every minute. When will you have the same number asyour friend? How much longer will it take you to finish your MnM's? At 10minutes you both will have 50 left. You will finish 1 min and 20 seconds after yourfriend.
Given:
The initial number of MNMs I have, x=200.
The number of MNM's eat by me every minute, p=15.
The initial number of MNMs my friend have, y=300.
The number of MNM's eat by friend every minute, q=25.
Let n be the number of minutes after which both will have the same number of MNM. Then, the amount of MNM remaining with me after n minutes is,
[tex]x-pn[/tex]The amount of MNM remaining with my friend after n minutes is,
[tex]y-qn[/tex]Equate the above expressions and substitute the values to find the number of minutes n.
[tex]\begin{gathered} x-pn=y-qn \\ 200-15n=300-25n \\ 25n-15n=300-200 \\ 10n=100 \\ n=\frac{100}{10} \\ n=10 \end{gathered}[/tex]Therefore, I will have the same number as my friend after 10 minutes.
The number of minutes taken by me to finish 200 MNM's is,
[tex]\begin{gathered} m=\frac{x}{p} \\ =\frac{200}{15} \\ =13\frac{5}{15} \\ =13\frac{1}{3}\text{minutes} \\ =13\text{minute}+\frac{1}{3}\min utes\times\frac{60\text{ seconds}}{1\text{ minute}} \\ =13\text{ minutes +20 seconds} \end{gathered}[/tex]So, I will take 13 minutes 20 seconds to finish the MNM's.
The number of minutes taken by my friend to finish 300 MNM's is,
[tex]\begin{gathered} k=\frac{y}{q} \\ =\frac{300}{25} \\ =12\text{ minutes} \end{gathered}[/tex]So, the friend will take 12 minutes to finish the MNM's.
So, I will finish
A package of 8-count AA batteries costs $6.16. A package of 20-count AA batteries costs $15.60. Which statement about the unit prices
is true?
The 8-count pack of AA batteries has a lower unit price of $0.77 per battery.
The 20-count pack of AA batteries has a lower unit price of $0.77 per battery.
The 20-count pack of AA batteries has a lower unit price of $0.78 per battery.
Answer:
it would be within the range of $0.77 <x< $0.78
Find the domain of the graphed function.A. -4sxs 8B. X2-4C. x is all real numbers.D. -4sxs 9
The domain of a function is the set of values over the x-axis where it is defined on a coordinate plane.
From the image, notice that the given graph is defined whenever x is between -4 and 9. Therefore, the domain of the function is:
[tex]-4\le x\le9[/tex]An athlete runs at a speed of 9 miles per hour. If one lap is 349 yards, how many laps does he run in 22 minutes
An athlete run in 22 minutes is 19.232 laps
Given,
An athlete runs at a speed of 9 miles per hour.
and, If one lap is 349 yards.
To find the how many laps does he run in 22 minutes?
Now, According to the question:
Firstly, Convert the mph into yard per minute,
Remember that:
I mile = 1,760 yard
1hour = 60 minute
Convert the speed in miles/hour to yards/minute
9 [tex]\frac{miles}{hour}[/tex] = 9[tex]\frac{1760}{60}[/tex] = 264 yard/ min
We know that
The speed is equal to divide the distance by the time
Let
s → the speed
d → the distance in yards
t → the time in minutes
Using the formula :
Speed = distance/ time
Solve the distance:
d = speed x time
Speed = 264 yard/ minute
Time = 22 minute
Therefore,
Distance = 264 x 22
Distance = 5,808 yards
Divide the distance by 302 yards to find out the number of laps
= 5,808/ 302 = 19.232 laps
Hence, An athlete run in 22 minutes is 19.232 laps
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Which is an example of a survey?A- collecting the cholesterol readings of a group of elderly people in a small townB- interviewing college students to find what percentage expect a job immediately after graduationC- testing the effectiveness of a hair product by allowing one group to use it and comparing results against a control groupD- testing the effectiveness of a mouthwash by allowing one group to use it and comparing results with those of a group that doesn't use ot
Answer: B- interviewing college students to find what percentage expect a job immediately after graduation
Surveys are meant to collect data that can be used to analyze a population as ahwole. This option analyzes the perspective of college students as a whole, whereas option A only focuses on a minority group of elderly. Additionally, options C and D are moreso experiments and not surveys that can be applied on a larger scale.
A college conducted a survey of randomly selected freshmen about their choice of major. The table shows the results of the survey. KS
ONLY F is correct;
Here, we want to select the correct inference from the data presented
f) We want to comapre the number of English freshmen and the undecided
Both have a count of 50; we can see that these values are equal and thus, we conclude that these two are equal
This makes the inference correct
g) Here, we want to compare Education freshmen to science freshmen or others
From the question, the number of education freshmen is 60
The number of science or others is (30+25) = 55
The number for education is greater and not less
This makes this option or inference incorrect
h) Here, we want to comapre Business/Education and Science/Engineering
Business OR Education is = 45 + 60 = 105
Science OR Engineering is = 35 + 40 = 75
Business/Education is greater and this makes this option or inference wrong
j) Here, we want to compare Business and English
Business is 45
English is 50
We can see that English is greater and this makes the inference/option wrong
the square root of 31 is closer to which number? 6 or 5.
Answer:
6
Explanation:
First, we find the squares of 5 and 6.
[tex]\begin{gathered} 5^2=25 \\ 31-25=6 \end{gathered}[/tex][tex]\begin{gathered} 6^2=36 \\ 36-31=5 \end{gathered}[/tex]We conclude therefore that the square root of 31 is closer to 6 since it has a smaller difference.
can someone please show me if im correct because i got 12
Given the expression:
-3 + 15
Let's evaluate the expression.
Here, we have an addition operation.
To perform the operation, add -3 and 15.
Hence, we have:
-3 + 15 = 12
Therefore, the answer to the operation is 12.
ANSWER:
12
Graph two or more functions in the same family for which the parameter being changed is the slope, m. and is less than 0.Refer to the graph of f(x) = x + 2
We have the expression:
[tex]f(x)=x+2[/tex]If the slope is changing being less than 0, that is:
Advantage Cellular offers a monthly plan of $25 for 500 minutes. What is the cost per minute? Round to the nearest hundredths place.
The cost per minute is $0.05 and it is round off to the nearest hundredths place.
Round off:
Rounding off means a number is made simpler by keeping its value intact but closer to the next number. It is done for whole numbers, and for decimals at various places of hundreds, tens, tenths, etc.
Given,
Advantage Cellular offers a monthly plan of $25 for 500 minutes.
Here we need to find the cost per minute and we have also need to round off the result to the nearest hundredth place.
We know that,
500 minutes cost = $25
So, to calculate the price for one minute, then we have to divide the cost by the total number of minutes.
So, it can be written as,
=> 25 ÷ 500
So, the cost for one minute is.
=> 0.050
When we rounded to the nearest hundredths place.
The 5 in the hundredths place rounds down to 5, or stays the same, because the digit to the right in the thousandths place is 0.
Therefore, the cost per minute is $0.05.
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May I get help, I know I have to multiply the possibilities, but I keep getting stuck
First we obtain each probability
The land has no oil
is a 45% chance that the land has oli , then the chance that the land has not oil is 55%
55% can be represented like 0.55
then the probability to the land has no oil is 0.55
The test shows that there is no oil
Kit claims to have an 80% of idicating oil, then the percent that there is no oil is 20%
20% can be represented like 0.2
the tne probability to shows that theere is no oil is 0.2
Finally
Multiply the probabilities to find the probability that say the land has no oil and the test shows that there is no oil
[tex]0.55\times0.2=0.11[/tex]then irhg toption is B
This not a test btw ! But can you please help me with this !
Using elimination:
[tex]\begin{gathered} (A)-3(B)\colon \\ 6x+12x-3y+3y=-4-15 \\ 18x=-19 \end{gathered}[/tex]Therefore, the answer is:
B) Multiply A by 1 and B by -3
the digits 1through 6 are used for a set of locker codes. suppose the digits cannot repeat. find the number of possible two digit codes and three digit codes. describe any pattern and use it to predict the number of possible five digit codes
SOLUTION
This is a permutation problem.
a) To find the number of possible two digits codes
[tex]^6P_2[/tex][tex]^6P_2=\frac{6!}{(6-2)!}[/tex][tex]\begin{gathered} =\frac{6!}{4!} \\ =\frac{720}{24} \\ =30\text{ ways} \end{gathered}[/tex]There are 30 possible two-digit codes pattern.
b) To find the number of three digits codes
[tex]\begin{gathered} ^6P_3=\text{ }\frac{6!}{(6-3)!} \\ \text{ =}\frac{6!}{3!} \\ \text{ =}\frac{720}{6} \\ \text{ = 120 ways} \end{gathered}[/tex]There are 120 possible three-digit codes pattern.
Any other pattern can be calculated using
[tex]\begin{gathered} ^6P_r \\ \text{where r is the number of digits code (1,2,3,4,5,6)} \end{gathered}[/tex]So to predict the number of possible five-digit codes will be:
[tex]^6P_5[/tex][tex]\begin{gathered} =\frac{6!}{(6-5)!} \\ =\frac{6!}{1!} \\ =720\text{ways} \end{gathered}[/tex]There are 720 different possible five-digit codes
the answer of this question is 720 ways
What is the value of the expression 4x−y2y+x when x = 3 and y = 3? −31918
7 ( 1 + 3 )
Solve the sum inside the parentheses ( 1 + 3 = 4 )
7 ( 4 )
multiply
7*4 = 28
Since the sum must equal 28
7 + 21 = 28
Correct option = 7+21
You buy items costing $3000 and finance the cost with a simple interest fixed installment loan at 5% simple interest per year. The finance charge is $600.a) How many years will you be paying?b) What is your monthly payment?
Given:
The principal amount is P = $3000.
The rate of interest is r = 5% = 0.05.
The interest rate is A = $600.
The objective is,
a) To find the number of years.
b) To find the monthly payment.
Explanation:
a)
The general formula for simple interest is,
[tex]A=P\times n\times r\text{ . . . . . .(1)}[/tex]To find n:
On plugging the given values in equation (1),
[tex]\begin{gathered} 600=3000\times n\times0.05 \\ n=\frac{600}{3000\times0.05} \\ n=4 \end{gathered}[/tex]b)
Since, the total amount of the item can be calculated as,
[tex]T=A+P\text{ .. . . . (2)}[/tex]On plugging the obtained values in equation (2),
[tex]\begin{gathered} T=600+3000 \\ T=3600 \end{gathered}[/tex]To find monthly payment:
Now, the monthly payment can be calculated as,
[tex]m=\frac{T}{n\times12}\text{ . . . . .(3)}[/tex]Here, m represents the monthly payment, the product of 12 is used to convert the number of years into the number of months.
On plugging the obtained values in equation (3),
[tex]\begin{gathered} m=\frac{3600}{4\times12} \\ m=75 \end{gathered}[/tex]Hence,
a) The number of years is 4 years.
b) The monthly payment is $75.
I need help with my algebra
We have the next equation line:
[tex]3x-y\text{ = 5}[/tex]We need to solve the equation for y to get the equation form
[tex]-y\text{ =5-3x}[/tex]Multiply the equation by -1
[tex](-1)-y\text{ =(-1)(5-3x)}[/tex][tex]y\text{ = -5+3x}[/tex]Where the y-intercept is -5 and the slope is 3x.
To find the line parallel we need to know that the parallel lines have the same slope.
The parallel line also intercepts y at point (0,-7).
[tex]y=mx+b[/tex]Replace the slope=m = 3
and the y-intercept is -7.
So the parallel line is:
[tex]y=3x-7[/tex]Which equation, written in the form of y = x + b, represents the table of values?
Let:
[tex]\begin{gathered} (x1,y1)=(2,7) \\ (x2,y2)=(5,10) \end{gathered}[/tex][tex]\begin{gathered} x=2,y=7 \\ 7=2m+b \\ ---------------- \\ x=5,y=10 \\ 10=5m+b \\ ---------- \\ Let\colon \\ 2m+b=7_{\text{ }}(1) \\ 5m+b=10_{\text{ }}(2) \\ (2)-(1) \\ 5m-2m+b-b=10-7 \\ 3m=3 \\ m=1 \end{gathered}[/tex]Replace m into (1):
[tex]\begin{gathered} 2(1)+b=7 \\ 2+b=7 \\ b=7-2 \\ b=5 \end{gathered}[/tex]Answer:
[tex]y=x+5[/tex]Ashlee was born on 09/08/1981. How many eight digit codes could she make using the digits in her birthday
Ashlee could make 2520 eight digit codes using the digits in her birthday .
In the question
it is given that
the birthdate of Ashlee is 09/08/1981.
So, the number of digits are 0,0,1,1,8,8,9,9 .
number of digits = 8
So, 8 digits can be arranged in 8! ways
8! = 8*7*6*5*4*3*2*1 = 40320
Repeating digits are
0 repeated 2 times = 2! = 2
1 repeated 2 times = 2! = 2
8 repeated 2 times = 2! = 2
9 repeated 2 times = 2! =2
So, the number of 8 digits codes is = 8!/(2!*2!*2!*2!)
= 8!/(2*2*2*2)
= 40320/16
= 2520
Therefore , Ashlee could make 2520 eight digit codes using the digits in her birthday .
The given question is incomplete , the complete question is
Ashlee was born on 09/08/1981. How many eight digit codes could she make using the digits in her birthday ?
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Assume the random variablex is normally distributed with mean p=85 and standard deviation o=5. Find the indicated probabiliP73
Remember that
z =(x - μ)/σ
we have
μ=85
σ=5
For x=73
Find out the value of Z1
z1=(73-85)/5
z1=-2.4
For x=76
Find out the value of Z2
z2=(76-85)/5
z2=-1.8
using a z-scores table values
we have that
P(73On the desmos app can you have more standard forms or only one?
Answer: I am pretty sure you can only have one.
Step-by-step explanation:
what is the slope for (0,-3),(-3,2)
Given the general rule for the slope:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]We have the following in this case:
[tex]\begin{gathered} (x_1,y_1)=(0,-3) \\ (x_2,y_2)=(-3,2) \\ \Rightarrow m=\frac{2-(-3)}{-3-0}=\frac{2+3}{-3}=-\frac{5}{3} \\ m=-\frac{5}{3} \end{gathered}[/tex]therefore, the slope is m=-5/3
What is the day supply for 30 tablets if the direction is a half tabletsFor 7 days then 1 once daily
Answer:
33 days
Explanation:
If you use a half tablet for 7 days, you will consume:
0.5 x 7 days = 3.5 tablets.
So, at the end of the 7th day, you will consume 3.5 tablets and you will have 26.5 tablets left because:
30 tablets - 3.5 tablet = 26.5 tablets.
Then, you will consume 1 tablet daily, so if you have 26.5 tablets, you will have a tablet daily for 26 days.
Therefore, the day supply for 30 tablets will be the sum of the initial 7 days and the 26 days, so:
7 days + 26 days = 33 days.
Therefore, the answer is 33 days
Write the equation in standard form for the hyperbola with vertices (-9,0) and (9,0) and a conjugate axis of length 16
The given vertices are (-9,0) and (9,0).
Notice that they lie on the x-axis since they have 0 as their y-coordinate.
Hence, the hyperbola is a horizontal hyperbola.
Recall that the equation of a horizontal hyperbola is given as:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]Where (h,k) is the center and a>b.
As both vertices are equidistant from the origin, the center of the hyperbola is (0,0), and the equation becomes:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]Note that the vertices are at (-a,0) and (a,0).
Compare with the given vertices (-9,0) and (9,0). It follows that a=9.
Substitute this into the equation:
[tex]\frac{x^2}{9^2}-\frac{y^2}{b^2}=1[/tex]Recall that the length of the conjugate axis is given as 2b, it follows that:
[tex]\begin{gathered} 2b=16 \\ \Rightarrow b=\frac{16}{2}=8 \end{gathered}[/tex]Substitute b=8 into the equation:
[tex]\begin{gathered} \frac{x^2}{9^2}-\frac{y^2}{8^2}=1 \\ \Rightarrow\frac{x^2}{81}-\frac{y^2}{64}=1 \end{gathered}[/tex]The required equation in standard form is:
[tex]\frac{x^2}{81}-\frac{y^2}{64}=1[/tex]