Multiplying Polynomials. Find the product and write the answer in standard form.
Given:
There are given that the expression:
[tex]-9b(a+4b)[/tex]Explanation:
Multiply -9b into the value of bracket.
So,
[tex]-9b(a+4b)=-9ab-36b^2[/tex]Final answer:
Hence, the equation is shown below:
[tex]-9ab-36b^2[/tex]
estimate 794 divided by 18=?
Answer:
C 40
Step-by-step explanation:
794 is about 800
18 is about 20
800/20=40
Solve Each System by Elimination:-12x-2y=30-4x+y=-5
We solve as follows:
-12x - 2y = 30
2(-4x + y = -5)
---------------------
-12x -2y = 30
-8x + 2y = -10
--------------------
-20x = 20 => x = -1
Now we replace the value of x in one of the original equations to solve for y, that is:
-4(-1) + y = -5 => 4 + y = -5 => y = -9
So, the solution is the point (-1, -9).
You need to measure the depth of a large lake. Since the sonar equipment is very expensive, you decide to use your friend's boat. The boat has an anchor on a 100 ft line. You take the boat out to the middle of the lake on a
windy day. You drop the anchor and let the wind push the boat until the anchor line is tight. Your GPS tells you that the boat has moved 82 feet. Assuming the bottom of the lake is flat, what is the depth of the lake?
Using the Pythagorean Theorem, the depth of the lake is 57.24 feet.
What is the depth?According to the Pythagorean Theorem, the depth or height is the difference between the squared root of the hypothenuse and the base.
The Pythagorean Theorem Formula is as follows:
a² + b² = c²
Where:
a = side of the right triangle (height, depth, or perpendicular)
b = side of the right triangle (the base)
c = hypotenuse (the longest part or hypothenuse)
Therefore, the depth is:
a² = c² - b²
a² = 100² - 82²
a² = 10,000 - 6,724
a² = 3,276
a = √3,276
a = 57.24
= 57.24 feet
Thus, assuming a flat-bottom lake, its depth is 57.24 feet.
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well I'm stuck on this homework question and need help please thank you
Which of the following inequalities would have solutions of -1, 1, 3, 4?Mark all that apply.A e > -1Bf <6c d < 4Db> -1EC < 5Fa> 0
Notice that for option B
f< 6 means that all numbers less than 6 are solution to the inequality, also notice that -1,1,3 and 4 are less than 6.
An analogous reasoning apllies for option E, all numbers less than 5 are solution to the inequality c<5 then -1,1,3 and 4 are solution.
For the rest of the inequalities at least one of the provided numbers are no solution for the inequality.
9) Professor Elderman has given the same multiple-choice final exam in his Principles ofMicroeconomics class for many years. After examining his records from the past 10 years, hefinds that the scores have a mean of 76 and a standard deviation of 12.Professor Elderman offers his class of 36 a pizza party if the class average is above 80. What isthe probability that he will have to deliver on his promise?
In this case, we'll have to carry out several steps to find the solution.
Step 01:
data:
score:
mean = 76
standard deviation = 12
Step 02:
probability:
normal distribution:
normal distribution diagram:
mean + 1 standard deviation = 76 + 12 = 88
probability (class average score is above 80):
p (score > 80) = 34% + 13.5% + 2.4% + 0.1% = 50% = 0.5
The answer is:
p (score > 80) = 50% = 0.5
You want to hang a picture on your wall the wall is 20 feet long you want to have at least 3 feet on both sides of the painting for plans to be placed right a quality showing the possible lanes of you being mean
We can consider
x = length of the painting
20 ft wall long
we want at least 3 ft per side so the total available length is 20-6= 14
x must be shorter than 14
inequality
x< 20-2(3)
x<14
One-third of a number b multiplied by -11 is more than 3 2
y varies inversely as x. y = 18 when x = 7. Find y when x = 3.y=(Simplify your answer.)
Use four rectangles to estimate the area between the graph of the function f(x) = Ty and the taxis on the interval 12, 6) using the left endpointsof the subintervals as the sample points. Write the exact answer, Do not round,
To find the area using four rectangles, we will use the following equation:
[tex]Area\approx A_1_{}+A_2+A_3+A_4[/tex][tex]Area\approx f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x[/tex][tex]Area\approx f(3)\Delta x+f(4)\Delta x+f(5)\Delta x+f(6)\Delta x[/tex][tex]Area\approx(\frac{6}{7(3)})(1)+(\frac{6}{7(4)})(1)+(\frac{6}{7(5)})(1)+(\frac{6}{7(6)})(1)[/tex][tex]Area\approx\frac{57}{70}[/tex]Write an inequality for the graph shown below. Use x for your variable. + -11-10-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 X
according to the graph, inequality is:
[tex]x\ge3[/tex]Word Problems: Define your variable, write and solve the equation. 9. You have $75 to spend at the grocery store. You get $23 in change. How much money do you spend? Define your variable: Equation: Answer:
ANSWER
Variable : how much money was spent (x)
Equation: x + 23 = 75
Answer: $52
EXPLANATION
Let the variable (amount of money spent) be x.
Initially, you had $75 and you spend some money (x) and you are left with $23 in change.
This means that the sum of the money you spent and the change you received is $75.
That is:
x + 23 = 75
That is the equation.
To find how much money you spent, we have to find the value of x by collecting like terms:
=> x = 75 - 23
x = $52
You spent $52.
As I am completely brand new to this subject/branch of mathematics, please explain thoroughly, step by step on how to complete this This is a practice from my ACT prep guide take your time, as there is no rush *Ignore the last answer option
Remember that
The difference of squares is of the form
[tex](a+b)(a-b)=a^2-b^2[/tex]In this problem we have
[tex](3x-4y^2)(3x+4y^2)[/tex]so
a=3x
b=4y^2
therefore
Apply the difference of squares
[tex](3x-4y^2)(3x+4y^2)=(3x)^2-(4y^2)^2=9x^2-16y^4[/tex]Given parallelogram ABCD, diagonals AC and BD intersect at point E. AE=2x, BE=y+10, CE=x+2 and DE=4y−8. Find the length of AC.A. 8B. 6C. 2D. 4
Answer:
the length of the diagonal AC is;
[tex]8[/tex]Explanation:
Given the parallelogram ABCD, diagonals AC and BD intersect at point E.
[tex]\begin{gathered} AE=2x \\ CE=x+2 \\ BE=y+10 \\ DE=4y+8 \end{gathered}[/tex]Recall that the diagonals of a parallelogram bisect each other;
So;
[tex]AE=CE[/tex]substituting AE and CE;
[tex]\begin{gathered} 2x=x+2 \\ 2x-x=2 \\ x=2 \end{gathered}[/tex]To calculate the length of AC;
[tex]\begin{gathered} AC=2x+x+2=3x+2 \\ since\text{ x=2} \\ AC=3x+2=3(2)+2 \\ AC=6+2 \\ AC=8 \end{gathered}[/tex]Therefore, the length of the diagonal AC is;
[tex]8[/tex]Nathan and some friends are going to the movies. At the theater, they sell a bag of popcorn for $6 and a drink for $4. How much would it cost if they bought 5 bags of popcorn and 7 drinks? How much would it cost if they bought pp bags of popcorn and dd drinks?Total cost, 5 bags of popcorn, and 7 drinks: Total cost, p bags of popcorn and d drinks:
a) Since the cost of a bag of popcorn is $6 and the cost of a drink is $4,
[tex]\begin{gathered} T=6\cdot5+7\cdot4 \\ \Rightarrow T=30+28=58 \\ \Rightarrow T=58 \end{gathered}[/tex]Therefore, the answer to the first question is $58.
b) Substitute 5 for p and 7 for d in the expression above; therefore,
[tex]T=6p+4d[/tex]The total cost is given by the equation T=6p+4d, where T is in dollars, p is the number of bags of popcorn and d is the number of drinks.
XYZ is a right-angled triangle. A is a point on line XZ and B is a point on line XY. XA = XB What is the size of angle XAB? xzy= 34 degrees
A. Monkey man
Step-by-step explanation:
M+o+n+k+e
2 dot plots. Both number lines go from 0 to 10. Plot 1 is titled fifth grade. There are 2 dots above 1, 3 above 2, 1 above 3, 4 above 4, 5 above 5, 5 above 6, 2 above 7, 2 above 8, 0 above 9, 0 above 10. Plot 2 is titled seventh grade. There are 2 dots above 0, 2 above 1, 3 above 2, 5 above 3, 5 above 4, 3 above 5, 3 above 6, 1 above 7, and 0 above 8, 9, and 10.
The dot plot shows the number of hours, to the nearest hour, that a sample of 5th graders and 7th graders spend watching television each week. What are the mean and median?
The 5th-grade mean is
.
The 7th-grade mean is
.
The 5th-grade median is
.
The 7th-grade median is
.
The mean and the median for each data-set are given as follows:
5-th grade students:
Mean: 4.67Median: 5 hours.7-th grade students:
Mean: 3.46 hours.Median: 4 hours.Dot plotA dot plot shows the number of times that each observation appears on a data-set.
Hence the hours of the 5th-graders are as follows:
1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 8, 8
The mean is the sum of all the numbers of hours divided by the number of students, hence:
Mean = (2 x 1 + 3 x 2 + 1 x 3 + 4 x 4 + 5 x 5 + 5 x 6 + 2 x 7 + 2 x 8)/(2 + 3 + 1 + 4 + 5 + 5 + 2 + 2) = 4.67.
There are 24 elements in the data-set, hence the median is the mean of the 12th and the 13th element, as follows:
Median = (5 + 5)/2 = 5.
Hence the hours of the 7th-graders are as follows:
0,0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7.
Hence the mean is:
Mean = (2 x 0 + 2 x 1 + 3 x 2 + 5 x 3 + 5 x 4 + 3 x 5 + 3 x 6 + 1 x 7)/24 = 3.46.
The 12th element is of 3, the 13th of 5, hence the median is:
Median = (3 + 5)/2 = 4.
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In 1990, there were 1330 registered alpacas in the United States. By summer of 2000, there were 29,856. What was the percent of increase in registered alpacas?
answer - 214% increase
explanation
formula = big number - small number ÷ original number
29856 - 1330 = 28,526 ÷ 1330 = 21.45
21.45 to percent = 214% then if rounded to nearest
In the diagram below, DE is parallel to yy. What is the value of x? 110° A. 90 0 B. 120 O C. 110 O D. 70
Angle shown x is corresponding angle to 110 degree angle shown (from property of transversal cutting a pair of parallel lines).
hence
x = 110
A math teacher said that 18 out of 25 students passed the test. What percent of thestudents did NOT pass the test?C.72%A.18%D.28%B.82%
Let's begin by listing out the given information:
18 students passed
Total students = 25
Number of people who did not pass = 25 - 18 = 7
7 out of 25 students did not pass. This has its percentage as:
[tex]\begin{gathered} \frac{7}{25}\cdot100\text{\%}=28\text{\%} \\ \Rightarrow28\text{\%} \end{gathered}[/tex]Therefore, 28% of the students did not pass
Two train leave stations 210 miles apart at the same time and travel toward each other. One train travels at 80 miles per hour while the other traves a 70miles per hout. How long will it take for the two trains to meet?___ hours Do not do any rounding
SOLUTION
At the same time t,
Train 1 would have covered a distance of 80t, since distance = average speed x time.
Train 2 would have covered a distance of 70t.
Now both added should give 210 miles
That is 80t + 70t = 210
150t = 210
t = 210/150
t = 1.4 hours
Please see photo checking my work I think it is all the students attending the college
Answer:
A population is the entire group that you want to draw conclusions about.
So, in this exercise, the population is all the students attending the college.
Round 13.134 to the nearest tenth.
Answer: 13.1
13.1
Rounded to the nearest 0.1 or
the Tenths Place.
Explanation:
13.134
You rounded to the nearest tenths place. The 1 in the tenths place rounds down to 1, or stays the same, because the digit to the right in the hundredths place is 3.
13.1
When the digit to the right is less than 5 we round toward 0.
13.134 was rounded down toward zero to 13.1
x³ - 3x = 37
Help please :(
15. x=m<1=I'll upload a picture of my HW
Opposite angles are the same
Then:
[tex]\begin{gathered} 6x+4=8x\text{ - }18 \\ 18+4=8x\text{ -}6x \\ 22=\text{ 2x} \\ x=\text{ 22/2} \\ x=11 \end{gathered}[/tex]So: (Remember the line has a 180º degrees
a student teacher is given a guideline that a student should be able to finish a 32 question test in 28 minutes if the student teacher is planning to give a test that contains 160 questions and the average students complete question of the same rate as previously State how many minutes should you plan for the average student to complete the test.
since the student do 32 questions in a time of 28 minutes, then the rate of response is 32/28=8/7 questions/minute ,then the students will be able to complete 160 questions in 20(7)= 140 minutes
State 3 solutions to the inequality: (1 Point) 3−4>5
We have the inequality:
[tex]3x\text{ - 4 }>5[/tex]Let's find out 3 solutions, as follows:
3x - 4 > 5
Adding 4 at both sides of the inequality:
3x - 4 + 4 > 5 + 4
3x > 9
Dividing by 3 at both sides, we have:
3x/3 > 9/3
x > 3
Now, we can find 3 solutions that fulfill the condition of the inequality:
x = 4, x = 5, x = 6
These three solutions are > 3
5. (a) The table below shows the cumulative frequency distribution of the weight of 80 deer recorded by the zookeeper. Weight, w kg Cumulative Frequency 6 15 61-80 36 niger 81-100 58 y Determine the upper class boundary for the class 21-40. Determine the class width for the class 41-60. How many deer were recorded in the class 81-100. (iv) A deer was chosen at random from the 80 deer. What is the probability that the weight of the deer is more than 100.5 kg. Leave your answer as an EXACT value. [2]
STEP - BY - STEP EXPLANATION
What to find?
• The upper class boundary for the class 21 - 40
,• Class width for 41 - 60
,• The number of deer recorded in the class 81 - 100
Given:
(i) To find the class boundary for the class 21 - 40, we will first subtract 0.5 from 21 and then add 0.5 to 40.
That is;
20.5 - 40.5
Hence, the upper class boundary is 40.5
(ii) The class width for the class 41 - 60
The class width can be determine by subtracting 41 from 60.
That is;
[tex]60-41=19[/tex]Hence, class width = 19
(iii) Number of deer recorded in the class 81 - 100
This can be obtain by subtracting the cumulative frequency in the class from the cumulative frequency before it.
58 - 36 =22
Hence, we have 22 numbers of deer in the class 81 - 100.
(iv) A deer was chosen at random from the 80 deer. What is the probability that the weight of the deer is more than 100.5 kg.
We can solve this by first determining the number of deer that are above 100.5 kg.
Number of beer above 100.5 kg = 15 + 7 = 22
Total number of deer = 80
[tex]Probability=\frac{required\text{ outcome}}{all\text{ possible outcome}}[/tex][tex]=\frac{22}{80}[/tex][tex]=\frac{11}{40}[/tex]ANSWER
(i) 40.5
(ii) 19
(iii) 22
(iv) 11/40
En un depósito había 127 bolsas de harina  cada una de 60 kg se sacaron ocho camiones de 12 bolsas cada uno cuantos kilogramos de harina quedaron en el depósito 
Based on the number of bags of flour that were taken by the trucks and the number that were in the warehouse, the amount of kilograms left in the warehouse is 1,860 kg
How to find the number of kilograms?First, find the number of bags that were taken by the trucks by the formula:
= Number of trucks x Number of bags per truck
= 8 x 12
= 96 bags
This means that the number of bags left are:
= 127 bags - 96 bags taken
= 31 bags left
The number of kilograms of flour left is:
= Number of bags left x Number of kilograms per bag
= 31 x 60
= 1,860 kg
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