Answer:
a. Increasing
b. Increasing
c. Decreasing
Explanation:
Part A (g(x) = 4x – 2)
[tex]\begin{gathered} g(2)=4(2)-2=8-2=6 \\ g(1)=4(1)-2=4-2=2 \\ \frac{g(2)-g(1)}{2-1}=\frac{6-2}{1} \\ =\frac{4}{1} \\ =4 \end{gathered}[/tex]g(x) is increasing since its rate of change is positive.
Part B (f(x) = 3x)
[tex]\begin{gathered} f(2)=3(2)=6 \\ f(1)=3(1)=3 \\ \frac{f(2)-f(1)}{2-1}=\frac{6-3}{1} \\ =\frac{3}{1} \\ =3 \end{gathered}[/tex]f(x) is increasing since its rate of change is positive.
Part C (h(x) = 2-x)
[tex]\begin{gathered} h(2)=2-2=0 \\ h(1)=2-1=1 \\ \frac{h(2)-h(1)}{2-1}=\frac{0-1}{1} \\ =\frac{-1}{1} \\ =-1 \end{gathered}[/tex]h(x) is decreasing since its rate of change is negative.
A blood test is able yo detect the presence of the drug if there is at least 0.1 mg in your blood. How many days will it take before the test will come back negative? Explain your answer.
Mrs. Gomes found that 40% of students at her high school take chemistry. She randomly surveys 12 students. What is the probability that at most 4 students have taken chemistry? Round the answer to the nearest thousandth.READ ANSWERS!0.0080.4380.5620.665
If we don't allow to repeat a pick on the sample, than each student we pick will change the probability of picking the next one and, since we don't know the total number of students at her high school, we wouldn't be able to calculate this probability.
Thus, to calculate this with the provided information, we need to assume that we can repeat a student in the sample, that way the probability of picking a chemistry student will always be 40%.
So, with that assumption, we can use binomial probability to calculate the probability of picking at most from from that have taken chemistry on the 12 picked students.
Since we want at most 4, we have the consider the term of the binomial for which 0 students from chemistry were picked, the term for 1 from chesmitry, the term for 2, the term for 3 and the term for 4.
Let p be the probability of picking 1 student from chemistry, n be the total number of students picked and k be the number of students on those that are from chemistry. Then, each term will have the following:
[tex]B(k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}_{}[/tex]In this case, n is always 12 and p is always 0.4, so 1 - p is always 0.6.
k will vary from 0 to 4 and the total probability will be the sum of each term.
The term for 0 students from chemistry has k = 0, so:
[tex]B(0)=\frac{12!}{0!(12-0)!}(0.4)^0(0.6)^{12-0}=1\cdot1\cdot(0.6)^{12}=0.0021767\ldots[/tex]The term for 1 student from chemistry has k = 1, so:
[tex]B(1)=\frac{12!}{1!(12-1)!}(0.4)^1(0.6)^{12-1}=12\cdot0.4\cdot(0.6)^{11}=0.0174142\ldots[/tex]The term for 2 student from chemistry has k = 2, so:
[tex]B(2)=\frac{12!}{2!(12-2)!}(0.4)^2(0.6)^{12-2}=\frac{12\cdot11}{2}\cdot(0.4)^2\cdot(0.6)^{10}=0.0638522\ldots[/tex]The term for 3 student from chemistry has k = 3, so:
[tex]B(3)=\frac{12!}{3!(12-3)!}(0.4)^3(0.6)^{12-3}=\frac{12\cdot11\cdot10}{3\cdot2}\cdot(0.4)^3\cdot(0.6)^9=0.1418939\ldots[/tex]The term for 4 student from chemistry has k = 4, so:
[tex]B(4)=\frac{12!}{4!(12-4)!}(0.4)^4(0.6)^{12-4}=\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}\cdot(0.4)^4\cdot(0.6)^8=0.2128409\ldots[/tex]So, the total probability will be the sum of all these term:
[tex]\begin{gathered} P=B(0)+B(1)+B(2)+B(3)+B(4) \\ P=0.0021767\ldots+0.0174142\ldots+0.0638522\ldots+0.1418939\ldots+0.2128409\ldots \\ P=0.4381782\ldots\approx0.438 \end{gathered}[/tex]So, the probability is approximately 0.438.
Write an equation of a quadratic function that has x-intercepts -2 and -5.
We have to write an equation of a quadratic function that has x-intercepts -2 and -5. This is, a function that has two roots: -2, and -5.
For doing so, we remember that if a function has n roots, it can be written as:
[tex]f(x)=(x-x_1)(x-x_2)_{}\ldots(x-x_n_{})[/tex]In this case, replacing the roots we have, we obtain that a quadratic function that has x-intercepts -2 and -5 is:
[tex]\begin{gathered} f(x)=(x-(-2))(x-(-5)) \\ f(x)=(x+2)(x+5) \end{gathered}[/tex]Explain in 5-7 sentences what type of triangles can be used with the sine/cosine/tangent ratios.
The Solution.
The triangle that can be used with sine, cosine, and tangent ratios must have the following characteristics:
1. The triangle must be a right-angled triangle.
2. The triangle must have one of its angles as 90 degrees.
3. Another angle must be indicated as the angle of interest.
4. At least two sides and an angle must be identified/represented with values.
5. The ratios under consideration can be used to find all the angles and all the sides.
ABCD is a quadrilateral
Work out the length of CD
Give your answwer in 3 significant figures
The length of CD is 20.48 cm
In this question, we have been given a quadrilateral ABCD.
We need to find the length of CD.
From given figure, consider right triangle ABD
We find the value of sine of angle ADB
sin(∠ADB) = AB/DB
sin(38°) = 10 / DB
0.61587 = 10/DB
DB = 10/0.6157
DB = 16.24 cm
Applying sine rule to triangle BCD,
sin(B)/CD = sin(C)/DB
sin(105°) / CD = sin(50°) / 16.24
0.9659 / CD = 0.766 / 16.24
CD = (0.9659 * 16.24) / 0.766
CD = 20.48 cm
Therefore, the length of CD is 20.48 cm
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which point in the solution set of the given inequalities?
In the image, we can identify the region of the plane in which the two subregions overlap (the part of the graph which has vertical and horizontal lines). We can obtain the values of x and y in that region by means of the inequalities, in this way:
[tex]\begin{cases}x+y>2 \\ 4x+y\ge-1\end{cases}[/tex]We solve the first inequality for y
[tex]y>2-x[/tex]This gives us the first constrain for the value of y
Now, to obtain the second constrain we simply need to use the second inequality:
[tex]\begin{gathered} 4x+y\ge-1 \\ \Rightarrow y\ge-1-4x \end{gathered}[/tex]Finally, the last step is to get the constraints for the value of x.
For this, we can directly analyze the image of the region, notice how it covers any value of x. This means that there are no constraints for the value of x.
[tex]\Rightarrow x\in\mathfrak{\Re }[/tex]So, the region is given by:
[tex]x\in\mathfrak{\Re },\begin{cases}y>2-x \\ y\ge-4x-1\end{cases}[/tex]Finally, the point that is in the region is the one that satisfies the previous inequalities.
1. (2,0)
If x=2, then
[tex]\begin{gathered} ,\begin{cases}y>2-2=0 \\ y\ge-4(2)-1=-9\end{cases} \\ \Rightarrow y>0 \end{gathered}[/tex]So, (2,0) cannot be the correct option
2. (0,2)
[tex]\begin{gathered} \begin{cases}y>2-0=2 \\ y\ge-4(0)-1=-1\end{cases} \\ \Rightarrow y>2 \end{gathered}[/tex]This means that (0,2) cannot be the right option
3. (0,3) notice that the same condition (y>2) holds in this situation too (since x=0)
But y=3>2. This point indeed satisfies the conditions! (0,3) is the answer
4. (0,-1) Notice that in this case y=-1 but it should happen that y>2. So this point cannot be in the region
How to make 16 using the numbers 4,3,3,1
Answer:
[tex](4*3)+3+1[/tex]
[tex]4*3=12[/tex]
[tex]12+3+1=16[/tex]
Please help me with this pls.The question is:“Choose a phrase that describes the slope of each graph”
The Solution:
The slope of the graph is undefined.
Explanation:
The slope of the grap
f(x) = 4x2 + 5.0 - 3g(x) = 4.3 - 3x2 + 5Find (f +9)(x).O A. (f +g)(x) = 4x3 + x2 + 5x + 2O B. (f +g)(x) = 4x3 + 4x2 + 2x + 2O c. (f+g)(x) = 8x3 + 2x + 2O D. (f +g)(x) = -4x3 + 7x2 + 53 - 8SUBMIT
ANSWER
A. (f + g)(x) = 4x³ + x² + 5x + 2
EXPLANATION
To add this functions we have to combine and add like terms:
[tex]\begin{gathered} (f+g)(x)=f(x)+g(x) \\ (f+g)(x)=4x^2+5x-3+4x^3-3x^2+5 \\ (f+g)(x)=(4-3)x^2+5x+4x^3+(5-3) \\ (f+g)(x)=x^2+5x+4x^3+2 \\ \text{ordering the terms} \\ (f+g)(x)=4x^3+x^2+5x+2 \end{gathered}[/tex]What is the measure of z?zХГУ49Z =z = [?]VGive your answer in simplest form.Enter
In order to find the value of z, proceed as follow:
By using the Pythagorean theorem for the bigger triangle, you obtain:
[tex]\begin{gathered} z^2=(4+9)^2-x^2 \\ z^2=13^2-x^2 \end{gathered}[/tex]Now, if you consider the second bigger triangle:
[tex]y^2=z^2-9^2[/tex]And in the smaller triangle:
[tex]x^2=y^2+4^2[/tex]Next, replace the second equation for y^2 into the previous one:
[tex]\begin{gathered} x^2=(z^2-9)+4^2^{} \\ x^2=z^2-9+4^2 \\ x^2=z^2-9+16=z^2+7 \end{gathered}[/tex]Now, replace the previous expression for x^2 into the first equation for z^2 and simplify:
[tex]\begin{gathered} z^2=13^2-(z^2+7) \\ z^2+z^2=13^2+7 \\ 2z^2=169+7 \\ z^2=\frac{176}{2} \\ z^2=88 \\ z=\sqrt[]{88}=\sqrt[]{4\cdot22}=2\sqrt[]{22} \end{gathered}[/tex]Hence, the value of z is 2√22
considering the changes in the bank account balance whose account balance change the most
Let's begin by listing out the information given to us:
Customer A = -$95.00
Customer B = -$75.00
Customer C = -$40.00
Customer D = -$92.00
Observing the figures above, we will observe that the greatest or biggest change is observed with Customer A. The account of Customer A decreased the most decreasing by $95; this was $3 more than Customer
Charles asked his teammates how many hours they practiced swimming during the week. He recorded his data in thetable below and used the shaded columns to calculate three sample means. What is the range of the values for thesample means?463Hours of Swim Practice53352447u ooo53234.6123.754.75
gAnswer
Range of the sample means = 3.34
Explanation
The range is defined as the difference between the highest and the lowest number. Mathematically,
Range = Highest number - Lowest number
So, the range of the sample means is the difference between the highest and the lowest sample means.
Range of the sample means =
(Sample mean with the highest value) - (Sample mean with the lowest value)
From the table, we can calculate the sample means. The sample mean is given as the sum of the variables divided by the number of variables.
Note that we are told that Charles used only the shaded columns to calculate three sample means. The shaded columns, grouped include
4, 3, 3. Mean = 3.33
8, 5, 7. Mean = 6.67
6, 4, 3. Mean = 4.33
5, 3, 6. Mean = 4.67
Sample mean with the highest value = 6.67
Sample mean with the lowest value = 3.33
Range of the sample means = 6.67 - 3.33 = 3.34
Hope this Helps!!!
Rewrite the equation below so that it does not have fractions2x/3 - 2 = 3/7Do not use decimals in your answer
The given equation is expressed as
[tex]\begin{gathered} \frac{2}{3}x\text{ -2=}\frac{3}{7} \\ \text{The lowest common factor is 21. We would multiply both sides of the equation by 21. } \\ \frac{2}{3}x\times21\text{ - 2}\times21=\frac{3}{7}\times21 \\ 14x-42=9 \\ \\ \\ \end{gathered}[/tex][tex]3x + 2 = 3[/tex]please help, thanks!
Solve
[tex]\begin{gathered} 3x+2=3 \\ \text{Subtract 2 from both sides of the equation} \\ 3x+2-2=3-2 \\ 3x=1 \\ \text{Next step, you divide both sides by 3} \\ \frac{3x}{3}=\frac{1}{3} \\ x=\frac{1}{3} \end{gathered}[/tex]The answer is x = 1/3
use the Table and example((worth 20 points very easy))Holly is filing as a single taxpayer with a taxable income of $40,300.Find her federal taxes due.
We have to find the federal taxes that should pay Holly, a single taxpayer.
In the single table, we see that the taxes due will be given by
[tex]4,453.5+\frac{22}{100}\text{ of the amount above \$}38,700[/tex]The amount is $40,300, and applying the formula we obtain:
[tex]\begin{gathered} 4,453.5+\frac{22}{100}(40,300) \\ =4,453.5+8,866 \\ =13319.5 \end{gathered}[/tex]This means that the federal taxes due of Holly is $13,319.5.
A bin size ofis most appropriate for this set of data:1486, 1228, 1625, 1340, 1845, 1573, 1608, 1821, 1398, 1572, 1401, 1563,1482, 1374
GIVEN:
We are given a set of data as indicated in the attached image.
Required;
Determine the appropriate bin size for the given set of data.
Step-by-step solution;
A bin size is basically a class interval in statistics. Its a way of sorting data in a histogram, more like categories.
We start by identifying the smallest and largest data points and these are;
[tex]\begin{gathered} Least=1228 \\ \\ Greatest=1845 \end{gathered}[/tex]Now we determine the range of the data set,
[tex]\begin{gathered} Range=Greatest\text{ }value-Least\text{ }value \\ \\ Range=1845-1228 \\ \\ Range=617 \end{gathered}[/tex]Let us now choose 5 bins.
Next step, we divide the range by the bin size and we have;
[tex]\frac{617}{5}=123.4[/tex]We now have a class interval of 123.4
We can round this down to 100. That means an interval of 100 between the lower limit and the upper limit of each class interval.
Therefore,
ANSWER;
Option D; 100
For which equation would x = 4 be a solution?28 – 5.25 x = 2.754.25 x + 7 = 244.25 x ÷ 8 = 97 + 3.25 x = 29
Given
The equations,
a) 28 – 5.25 x = 2.75
b) 4.25 x + 7 = 24
c) 4.25 x ÷ 8 = 9
d) 7 + 3.25 x = 29.
To find:
For which equation would x = 4 be a solution?
Explanation:
It is given that,
a)
[tex]28-5.25x=2.75[/tex]Put x=4,
[tex]\begin{gathered} \Rightarrow28-5.25(4)=2.75 \\ \Rightarrow28-21=2.75 \\ \Rightarrow7=2.75,\text{ which is a contradiction.} \end{gathered}[/tex]Hence 4 is not a solution of a) 28 – 5.25 x = 2.75.
b)
[tex]4.25x+7=24[/tex]Put x=4,
[tex]\begin{gathered} \Rightarrow4.25(4)+7=24 \\ \Rightarrow17+7=24 \\ \Rightarrow24=24 \end{gathered}[/tex]Hence, 4 is the solution of b) 4.25 x + 7 = 24.
Thus, the answer is option b).
Val is going to plant y vegetable seeds in one garden and 4y +9 vegetable seeds in another. How many seeds is Val going to plant?
Given:
Val is going to plant y vegetable seeds in one garden and 4y +9 vegetable seeds in another.
So, the total seeds =
[tex]y+(4y+9)=y+4y+9=5y+9[/tex]How many different committees of six people can be selected from nine people?
ANSWER
[tex]84[/tex]EXPLANATION
To find the number of committees of six people that can be selected from nine people, we apply Combination:
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]Therefore, we have:
[tex]\begin{gathered} ^9C_6=\frac{9!}{(9-6)!6!} \\ \Rightarrow\frac{9\cdot8\cdot7\cdot6!}{3!6!} \\ \Rightarrow\frac{9\cdot8\cdot7}{3\cdot2\cdot1} \\ \Rightarrow84 \end{gathered}[/tex]Hence, there are 84 different ways of selecting.
Thomas is saving money for a few montain bike. the amount(a) Thomas needs to save is more than $50.45. which inequality models the amount Thomas needs to save A.$50.45=a. B.$60.45$50.45. D.a
From the information given in the statement you know that:
*a is the amount of money Thomas needs to save
*Thomas needs to save more than $ 50.45.
Then, the inequality that models the amount Thomas needs to save is
[tex]a>\text{ \$}50.45[/tex]use long or synthetic division to find the following quotients (w^3-2w^2-2w+1)÷(w-1)
The given polynimial exression is
(w^3-2w^2-2w+1)÷(w-1)
We wuld apply the long division method as shown below
The first step is todivide w^3 by w to give us w^2. w^2 is used to multiply w + 1 to give us w^3 + w^2 and this is subtracted from w^3 - 2w^2. The process continues until we cannot divide further. Thus, from the daigram above, the solution is
w^2 - 3w + 1
Given the function f(x)=x^2-5x-2, determine the average rate of change of the function over the interval -1≤x≤5
ANSWER
[tex]-1[/tex]EXPLANATION
Given;
[tex]\begin{gathered} f(x)=x^2-5x-2 \\ (-1,5) \end{gathered}[/tex][tex]\begin{gathered} f(-1)=(-1)^2-5(-1)-2 \\ =1+5-2 \\ =4 \end{gathered}[/tex]For x = 5;
[tex]\begin{gathered} f(5)=(5)^2-5(5)-2 \\ =25-25-2 \\ =-2 \end{gathered}[/tex]The average is calculated using;
[tex]\begin{gathered} \frac{f(-1)-f(5)}{-1-5} \\ =\frac{4-(-2)}{-1-5} \\ =\frac{6}{-6} \\ =-1 \end{gathered}[/tex]A
Describe the end behavior f(x)=x^3-4x^2+7
The behavior of the function falls to the left and rises to the right.
Given function f(x)=x^3-4x^2+7,
Firstly identify the degree of the function which is :
3
Now identify the leading coefficient which is :
1
Now, we know that if the degree is odd then the function ends in opposite direction.
And also the leading coefficient is positive which leads to rise of the graph to the right.
By using the degree and leading coefficient we can determine the behavior of the function by using below statements ;
1. Even and Positive: Rises to the left and rises to the right.
2. Even and Negative: Falls to the left and falls to the right.
3. Odd and Positive: Falls to the left and rises to the right.
4. Odd and Negative: Rises to the left and falls to the right
So the behavior of the function falls to the left and rises to the right.
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What is the volume of a figure with that is 5 inches wide, 10 inches tall and 5 inches long
The volume of a box is:
[tex]\begin{gathered} V=W\cdot T\cdot L \\ \end{gathered}[/tex]So for our problem will be:
[tex]V=5\cdot10\cdot5=250[/tex]Solve for q. 4(q + 17) − 12 = 0 q =
Answer:
q = -14
Step-by-step explanation:
4(q + 17) − 12 = 0
4q + 68 - 12 = 0
4q + 56 = 0
4(q + 14) = 0
4 = 0, q + 14 = 0
q = -14
I hope this helps!
Write an equation in slope-intercept form for the line described. Write the slope and y-intercept as improper fractions, if necessary.
passes through (−2, 2), perpendicular to y=−5x−8
Equation in slope-intercept form for the required line which passes through (-2, 2) and perpendicular to y= -5x-8 is equal to y = 5x+12.
As given in the question,
Line passes through the point ( -2, 2)
Required line is perpendicular to the given line y = -5x -8.
Let the point by ( x₁ , y₁) = ( -2, 2)
Standard equation of line is
y =mx +c
Slope = m
Slope of given line is -5
Lines are perpendicular
Slope of required line is 5
Required equation in slope-intercept form is:
y -2 =5(x-(-2))
⇒ y -2 = 5x +10
⇒ y= 5x +12
Therefore, equation in slope-intercept form for the required line which passes through (-2, 2) and perpendicular to y= -5x-8 is equal to
y = 5x+12.
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which best describes the solutions for this system? y=3x-4 2y=2x+1infinitely many solution not a solution there one solution??
The equations are y = 3x - 4 and 2y = 2x +1.
If the graph of equation intersect each other at one point then there is one solution, if line lies on one another then there are infinitely many solution and if lines are parallel to each other then there is no solution.
Plot the equation on the graph.
From the graph, it can be observed the lines intersect each other at one point only lies in first quadrant, so there is one solution located in the first quadrant.
Determine the number of significant figures in the following masses.
From the question;
We are to determine the number of significant figures in
a) 0.00040230g
b) 405,000g
a) For 0.00040230g
Trailling zeros to the right of the decimal point are significant
The digits zeros before 4 are not significant
Therefore, the mass has 5 significant figures
b) For 405,000g
Trailling zeros in a whole number with no decimal point are not significant
Therefore
The mass has 3 significant figures
For each equation determines whether y is a funcation of x
All linear relations are functions
The quadratic equations can be a function if we restrict the domain of it
The given relations are
[tex]x=-9y[/tex]This is a linear relation, then it is a function, we can put it in the form
[tex]y=-\frac{1}{9}x[/tex][tex]y=8x[/tex]This is a linear relation, then it is a function
[tex]x+3y=6[/tex]This is a linear relation, then it is a function, we can put it in the form
[tex]y=\frac{6-x}{3}[/tex][tex]x=2y^2-3[/tex]We will put it in the form of y as a subject
[tex]y^2=\frac{x+3}{2}[/tex]To find y we will take square root for both sides
[tex]\begin{gathered} \sqrt[]{y^2}=\pm\sqrt[]{\frac{x+3}{2}} \\ y=\pm\sqrt[]{\frac{x+3}{2}} \end{gathered}[/tex]That means each value of x has 2 values of y, then
This can not be a function
First rewrite 10/21 and 4/9 so that they have a common denominator
Answer:
10/21=30/63
4/9=28/63
Step-by-step explanation:
Find the lcm (least common multiple) of 21 an 9, which is 63. Then divide 21 and 9 separately by 63. The multiply that number by the numerator.