Let the output is y and the input is x, then
output is means y =
9 less than 5 times input means 9 less than 5x
Then
y = 5x - 9
What is the standard form of the complex number that point A represents?
Answer
-3 + 4i
Explanation
The standard form for a complex number is given by:
[tex]\begin{gathered} Z=a+bi \\ \text{Where:} \\ a\text{ is the real part,} \\ b\text{ is the imaginary part} \end{gathered}[/tex]From the graph, the coordinates of A corresponding to the real axis and imaginary axis is traced in blue color in the graph below:
Hence, the standard form of the complex number that a represents is: -3 + 4i
Khalil has 2 1/2 hours to finish 3 assignments if he divides his time evenly , how many hours can he give to each
In order to determine the time Khalil can give to each assignment, just divide the total time 2 1/2 between 3 as follow:
Write the mixed number as a fraction:
[tex]2\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}[/tex]Next, divide the previous result by 3:
[tex]\frac{\frac{5}{2}}{\frac{3}{1}}=\frac{5\cdot1}{2\cdot3}=\frac{5}{6}[/tex]Hence, the time Khalil can give to each assignment is 5/6 of an hour.
Find the measure of the indicated angle to the nearest degree.A. 63B. 25C. 31D. 27
The point of the problem is to remember the cosine relation. It says, in this case, that
[tex]\cos (?)=\frac{\text{adjacent side}}{Hypotenuse}\Rightarrow\begin{cases}\text{adjacent side}=6 \\ \text{Hypotenuse}=13\end{cases}\Rightarrow\cos (?)=\frac{6}{13}[/tex]Converting the last equation by the inverse function, we get
[tex]?=\cos ^{-1}(\frac{6}{13})\approx62.5[/tex]For the first decimal place (5) equals 5, and by the rounding rule to the nearest degree, we get 63. The answer is A.
Instructions: Factor 2x2 + 252 + 50. Rewrite the trinomial with the c-term expanded, using the two factors. Answer: 24 50
Given the polynomial:
[tex]undefined[/tex]Write the sequence {15, 31, 47, 63...} as a function A. A(n) = 16(n-1)B. A(n) = 15 + 16nC. A(n) = 15 + 16(n-1)D. 16n
To find the answer, we need to prove for every sequence as:
Answer A.
If n=1 then:
A(1) = 16(1-1) = 16*0 = 0
Since 0 is not in the sequence so, this is not the answer
Answer B.
If n=1 then:
A(1) = 15 + 16*1 = 31
Since 31 is not the first number of the sequence, this is not the answer
Answer D.
If n=1 then:
16n = 16*1 = 16
Since 16 is not in the sequence so, this is not the answer
Answer C.
If n = 1 then:
A(1) = 15 + 16(1-1) = 15
A(2) = 15 + 16(2-1) = 31
A(3) = 15 + 16(3-1) = 47
A(4) = 15 + 16(4-1) = 63
So, the answer is C
Answer: C. A(n) = 15 + 16(n-1)
Function f is defined by f(x) = 2x – 7 and g is defined by g(x) = 5*
Answer
f(3) = -1, f(2) = -3, f(1) = -5, f(0) = -7, f(-1) = -9
g(3) = 125, g(2) = 25, g(1) = 5, g(0) = 1, g(-1) = 0.2
Step-by-step explanation:
Given the following functions
f(x) =2x - 7
g(x) = 5^x
find f(3), f(2), f(1), f(0), and f(-1)
for the first function
f(x) = 2x - 7
f(3) means substitute x = 3 into the function
f(3) = 2(3) - 7
f(3) = 6 - 7
f(3) =-1
f(2), let x = 2
f(2) = 2(2) - 7
f(2) = 4 - 7
f(2) =-3
f(1) = 2(1) - 7
f(1) = 2 - 7
f(1) =-5
f(0) = 2(0) - 7
f(0) =0 - 7
f(0) = -7
f(-1) = 2(-1) - 7
f(-1) = -2 - 7
f(-1) = -9
g(x) = 5^x
find g(3), g(2), g(1), g(0), and g(-1)
g(3), substitute x = 3
g(3) = 5^3
g(3) = 5 x 5 x 5
g(3) = 125
g(2) = 5^2
g(2) = 5 x 5
g(2) = 25
g(1) = 5^1
g(1) = 5
g(0) = 5^0
any number raised to the power of zero = 1
g(0) = 1
g(-1) = 5^-1
g(-1) = 1/5
g(-1) = 0.2
2.3 I can apply the Pythagorean Theorem and Triangle Inequality.Which of the following could be lengths for a triangle?Show your work on a separate piece of paper.(Select all that apply.)5, 6, 9D 4,8, 127, 8, 17Are any of the selected triangles above right triangles?How do you know?Suami
For the triangle with sides 5, 6 and 9, you have:
5 + 6 > 9
5 + 9 > 6
9 + 6 > 9
9² ≠ 5² + 6²
≠ 25 + 36
≠ 61
Then, it is not a right triangle
For the triangle with sides 4, 8 and 12:
4 + 8 ≥ 12
in this case the triangle inequality is not present
12² ≠ 4² + 8²
Then, it is not a right triangle
For the triangle with sides 7, 8 and 17:
7 + 8 < 17
in this case the triangle inequality is not present
17² ≠ 8² + 7²
Then, it is not a right triangle
1. How much less is the area of a rectangular field 60 by 20
meters than that of a square field with the same perimeter?
The area of the rectangular field is 400m² less than the area of the square field.
How to find the area of a rectangle and square?A rectangle is a quadrilateral that has opposite sides equal to each other. Opposite side are also parallel to each other.
A square is a quadrilateral that has all sides equal to each other.
Therefore,
area of the rectangular field = lw
where
l = lengthw = widthTherefore,
area of the rectangular field = 60 × 20
area of the rectangular field = 1200 m²
The square field have the same perimeter with the rectangular field.
Hence,
perimeter of the rectangular field = 2(60 + 20)
perimeter of the rectangular field = 2(80)
perimeter of the rectangular field = 160 meters
Therefore,
perimeter of the square field = 4l
160 = 4l
l = 160 / 4
l = 40
Hence,
area of the square field = 40²
area of the square field = 1600 m²
Difference in area = 1600 - 1200
Difference in area = 400 m²
Therefore, the area of the square field is 400 metre square greater than the rectangular field.
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A cylinder truck all paint cans to be inches across the top diameter in about 10 inches high. How many cubic inches of pink it all to the nearest hundredth?
Given:
A cylinder truck all paint cans to be inches across the top diameter in about 10 inches high.
[tex]\begin{gathered} r=1.5in \\ h=10in \end{gathered}[/tex]Required:
To find the volume of the cylinder.
Explanation:
The volume of the cylinder is,
[tex]V=\pi r^2h[/tex]Therefore,
[tex]\begin{gathered} V=3.14\times1.5^2\times10 \\ \\ =3.14\times2.25\times10 \\ \\ =70.65in^3 \end{gathered}[/tex]Final Answer:
70.65 cubic inches of paint it hold.
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Given the table in I which represents function I.
x y
0 5
1 10
2 15
3 20
4 25
• Graph II shows Item II which represents the second function.
Let's determine the increasing and decreasing function.
For Item I, we can see that as the values of x increase, the values of y also increase. Since one variable increases as the other increases, the function in item I is increasing.
For the graph which shows item II, as the values of x increase, the values of y decrease, Since one variable decreases as the other variable decreases, the function in item I is decreasing.
Therefore, the function in item I is increasing, and the function in item II is decreasing.
ANSWER:
A. The function in item I is increasing, and the function in item II is decreasing.
3 2 — · — = _____ 8 5 2 9· — = _____ 3 7 8 — · — = _____ 8 7 x — · y = _____ y a b —— · — = _____ 2b c m n2 —- · —— = _____ 3n mGive the product in simplest form: 1 2 · 2— = _____ 2Give the product in simplest form: 1 2 — · 3 = _____ 4 Give the product in simplest form: 1 1 1— · 1— = _____ 2 2 Give the product in simplest form: 1 2 3— · 2— = _____ 4 3
Given:
[tex]\frac{3}{8}\cdot\frac{2}{5}[/tex]Required:
We need to multiply the given rational numbers.
Explanation:
Cancel out the common terms.
[tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{4}\cdot\frac{1}{5}[/tex][tex]Use\text{ }\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}.[/tex][tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{20}[/tex]Consider the number.
[tex]\frac{7}{8}\cdot\frac{8}{7}=\frac{1}{1}\cdot\frac{1}{1}[/tex]Cancel out the common multiples
[tex]9\cdot\frac{2}{3}[/tex][tex]9\cdot\frac{2}{3}=3\cdot2=6[/tex]Consider the number
[tex]\frac{7}{8}\cdot\frac{8}{7}[/tex]Cancel out the common multiples.
[tex]\frac{7}{8}\cdot\frac{8}{7}=\frac{1}{1}\cdot\frac{1}{1}[/tex][tex]\frac{7}{8}\cdot\frac{8}{7}=1[/tex]Consider the number
[tex]\frac{x}{y}\cdot y=x[/tex][tex]\frac{a}{2b}\cdot\frac{b}{c}=\frac{a}{2}\cdot\frac{1}{c}=\frac{a}{2c}[/tex][tex]\frac{m}{3n}\cdot\frac{n^2}{m}=\frac{1}{3}\cdot\frac{n}{m}=\frac{n}{3m}[/tex]Final answer:
[tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{20}[/tex][tex]9\cdot\frac{2}{3}=6[/tex][tex]\frac{7}{8}\cdot\frac{8}{7}=1[/tex][tex]\frac{x}{y}\cdot y=x[/tex][tex]\frac{a}{2b}\cdot\frac{b}{c}=\frac{a}{2c}[/tex][tex]\frac{m}{3n}\cdot\frac{n^2}{m}=\frac{n}{3m}[/tex]A length of 48 ft. gave Malama an area
of 96 sq. ft. What other length would
give her the same area (96 sq. ft.)?
4
Which graph represents the function over the interval [−3, 3]?f(x)=⌊x⌋−2
Given:
[tex]f(x)=x-2\text{ ,\lbrack-3,3\rbrack}[/tex]In ATUV, the measure of ZV=90°, TV = 28, UT = 53, and VU = 45. What ratiorepresents the cosecant of ZU?
cosecant = hypotenuse / opposite side
hypotenuse = 53
opposite side = 28
cosecant U = 53/28
In the picture, the first answer circled is the original answer of the problem. My math teacher simplified this to get the second circled answer. Could you explain how he simplified it?
We have an algebraic problem where we have to solve for "w"
[tex]3x+2k=\frac{15y}{9w-18v}[/tex]Solving for "w"
[tex]\begin{gathered} 9w-18v=\frac{15y}{3x+2k} \\ w=\frac{\frac{15y}{3x+2k}}{9}+\frac{18v}{9} \\ w=\frac{15y}{27x+18k}+2v \end{gathered}[/tex]The previous result is the solution to the problem without simplifying, the error is that you have in the image, in the denominator the factor "23x" in reality this is "27x"
Now we can simplify this by taking out the third part of the whole fractional term
For him we divide everything by 3, being the third part of 15, 27, and 18 respectively 5, 9, and 6.
[tex]w=\frac{5y}{9x+6k}+2v[/tex]Two students measured a box in class. They used a digital scale and found that the mass was 400 grams. They then measured the box found the length is 2cm, the width is 2cm, and the height is 1cm. What is the density of the object
Explanation
Step 1
the density of an object is given by:
[tex]\begin{gathered} density=\frac{mass_{object}}{volume_{object}} \\ \end{gathered}[/tex]Let
mass: 400 grams
length's box=2 cm
width´s box= 2 cm
height's box= 1 cm
Step 2
find the volume of the box
[tex]\begin{gathered} \text{Volume}=\text{ length}\cdot width\cdot height \\ \text{replacing} \\ \text{Volume= 2 cm }\cdot\text{ 2 cm }\cdot\text{ 1 cm} \\ \text{Volume}=\text{ 4 cubic cm} \end{gathered}[/tex]Step 3
finally, replace the values of mass and volume in the density equation
[tex]\begin{gathered} density=\frac{mass_{object}}{volume_{object}} \\ density=\frac{400\text{ grm}}{4cm^3} \\ \text{density}=100\frac{gr}{cm^3} \end{gathered}[/tex]I hope this helps you
Which family spends the largest dollar amount on transportation?Family AFamily BFamily C
SOLUTION:
Step 1:
In this question, we are given the following:
Which family spends the largest dollar amount on transportation?
a) Family A
b) Family B
c) Family C
Step 2:
The details of the solution are as follows:
a) Family A
Total income $ 5, 400
Amount spent on Transportation =
[tex]\begin{gathered} 11\text{ \% of \$ 5,400} \\ \frac{11}{100}\text{ x \$ 5, 400} \\ =\text{ }\frac{59400}{100} \\ =\text{ 594} \\ =\text{ \$ 594} \\ So,\text{ Family A spent \$ 594 on Transportation} \end{gathered}[/tex]b) Family B
Total income $ 4,675
Amount spent on Transportation =
+
[tex]\begin{gathered} 13\text{\% of \$ }4,675 \\ \frac{13}{100}\text{ x \$ }4,675 \\ =\text{ }\frac{60775}{100} \\ =607.75 \\ =\text{ }607.\text{ 75 dollars} \\ So,\text{ Family B spent \$ 607.75 on Transportation} \end{gathered}[/tex]c) Family C:
Total income $ 6,675
Amount spent on Transportation =
+
[tex]\begin{gathered} 9\text{\% of \$ }6,675 \\ \frac{9}{100}\text{ x \$ }6,675 \\ =\text{ }\frac{60,075}{100} \\ =600.75 \\ =\text{ }600.75\text{ dollars} \\ So,\text{ Family C spent \$ 600.75 on Transportation} \end{gathered}[/tex]CONCLUSION:
From the above analysis,
we can see that Family B spent the largest dollar amount on Transportation with the sum of $ 607. 75 ( which is 13% of $ 4,675)
Janelle is conducting an experiment to determine whether a new medication is effective in reducing sneezing. She finds 1,000 volunteers with sneezing issues and divides them into two groups. The control group does not receive any medication; the treatment group receives the medication. The patients in the treatment group show reduced signs of sneezing. What can Janelle conclude from this experiment?
Answer:
Step-by-step explanation:
Miscavage Corporation has two divisions: the Beta Division and the Alpha Division.
The Beta Division has:
sales of $320,000,
variable expenses of $158,100,
and traceable fixed expenses of $72,300.
The Alpha Division has:
sales of $630,000,
variable expenses of $343,800,
and traceable fixed expenses of $135,100.
The total amount of common fixed expenses not traceable to the individual divisions is $137,200.
What is the total company's net operating income?
The total net operating income (NOI) of both divisions is $1,03,500.
What is net operating income?Real estate professionals use the formula known as Net Operating Income, or NOI, to quickly determine the profitability of a specific investment. After deducting required operating costs, NOI calculates the revenue and profitability of investment real estate property. Let's say, for illustration purposes, that you own a duplex with a gross monthly income of $2,000 and monthly operating expenses of $400. You would start with your annual gross income ($24,000) and deduct your operating expenses ($4,800) to arrive at your net operating income.So, the total net operating income:
The formula for net operating income: NOI = Gross Income - Operating ExpensesNow, substitute the values and get the NOI as follows:
NOI = Gross Income - Operating ExpensesNOI = (Sales+Sales) - [(variable expenses + variable expenses) + (fixed expenses + fixed expenses) + 137,200] NOI = (320,000+630,000) - [(158,100 + 343,800) + (72,300 + 135,100) + 137,200]NOI = 9,50,000 - (5,01,900 + 2,07,400 + 137,200)NOI = 9,50,000 - 8,46,500NOI = 1,03,500Therefore, the total net operating income (NOI) of both divisions is $1,03,500.
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French cooks usually weigh ingredients. A French recipe uses 225 grams of granulated sugar.How many cups are needed if there are 2 cups of sugar per pound: (Note that you are changingfrom units of weight, grams, to units of volume, cups. There are 453.5 grams/pound)cups
Given:
The amount of granulated sugar used fo French fries, x=225 g.
1 pound=2cups.
1 pound=453.5g.
Since 1 pound =453.5 g,
[tex]1\text{ g=}\frac{1}{453.5}\text{ pound}[/tex]Therefore, 225 grams can be expressed in pound as,
[tex]\begin{gathered} 225\text{ g=}225\text{ g}\times\frac{\frac{1}{453.5}pound}{1\text{ g}} \\ =\frac{225}{453.5}pound \\ \cong0.496\text{ pound} \end{gathered}[/tex]Since 1 pound =2 cups, we can write
[tex]\begin{gathered} 0.496pound=0.496pound\times\frac{2cup}{1\text{ pound}} \\ =0.992\text{ cups} \\ \cong1\text{ cup} \end{gathered}[/tex]Therefore, 1 cup is needed.
the hypotenuse of a right triangle is 5 ft long. the shorter leg is 1 ft shorter than the longer leg. find the side lengths of the triangle
the hypotenuse of the right angle triangle is h = 5 ft
it is given that
the shorter leg is 1 ft shorter than the longer leg.
let the shorter leg is a and longer leg is b
the
b - a = 1
b = 1 + a
in the traingle using Pythagoras theorem,
[tex]a^2+b^2=h^2[/tex]put he values,
[tex]a^2+(1+a)^2=5^2[/tex][tex]\begin{gathered} a^2+1+a^2+2a=25 \\ 2a^2+2a-24=0 \\ a^2+a-12=0 \end{gathered}[/tex][tex]\begin{gathered} a^2+4a-3a-12=0_{} \\ a(a+4)-3(a+4)=0 \\ (a+4)(a-3)=0 \end{gathered}[/tex]a + 4 = 0
a = - 4
and
a - 3 = 0
a = 3
so, the longer leg is b = a + 1 = 3 + 1 = 4
thus, the answer is
shorter leg = 3 ft
longer length = 4 ft
hypotenuse = 5 ft
5. 8.G.1.5 Right triangle ABC and right triangle ACD overlap as shown below. Angle DAC measures 20° and angle BCA measures 30°. B D to 20- A 30° C not drawn to scale What are the values of and y?
In any triangle, the sum of the internal angle is always up to 180º.
Then, for triangle ABC:
[tex]90º+30º+(x+20º)=180º[/tex]Use it to solve x:
[tex]\begin{gathered} 90º+30º+x+20º=180º \\ x=180º-90º-30º-20º \\ x=40º \end{gathered}[/tex]In the triangle ACD:
[tex]90º+20º+(y+30º)=180º[/tex]Use it to solve y:
[tex]\begin{gathered} 90º+20º+y+30º=180º \\ y=180º-90º-20º-30º \\ y=40º \end{gathered}[/tex]Then, the value for x is 40º and the value for y is also 40ºThe coordinates of three vertices of a rectangle are (3,7), (-3,5), and (0,-4). What are the coordinates of the fourth vertex?A. (6,-2)B. (-2,6)C. (6,2)D. (-2,-6)
ANSWER
A. (6, -2)
EXPLANATION
Let's graph these three vertices,
The fourth vertex must be at the same distance from (0, -4) as vertex (3, 7) is from (-3, 5),
Note that the horizontal distance between these two points is 6 units and the vertical distance is 2 units. The fourth vertex is,
[tex](0+6,-4+2)=(6,-2)[/tex]Hence, the fourth vertex is (6, -2)
WhichIs 9.56556555... a rational or irrationalnumber? Highlight the correct answer below.181a)Whicha) Rational numberb) Irrational numberthat ap
Answer
Option B is correct.
9.56556555... is an irrational number.
Step-by-step Explanation
Rational numbers are numbers that can be expressed as a clear fraction consisting of the numerator and the denominator both being integers.
The decimal form or decimal expansion of a rational number terminates after a particular/finite number of digits (e.g., 0.25, 0.762 etc.) or begins to repeat/recur the same sequence over and over again (e.g., 0.333..., 0.267267... etc)
Anything other than these two rules, the number is regarded as an irrational number.
The number given is 9.56556555...
The dots indicate thst the numbers after the decimal point conbtinue till eternity.
Observing the numbers after the decimal point for the given number, one can see that 565 repeats once and then the number after the second 565 is 55, indicating that the 565 doesn't recur till infinity.
Since the numbers after the decimal point doesnt contain a finite number of digits and the numbers don't recur till infinity, we can conclude that 9.56556555... isn't a rational number.
Hope this Helps!!!
Macky Pangan invested ₱2,500 at the end of every 3-month period for 5 years, at 8% interest compounded quarterly. How much is Macky’s investment worth after 5 years?
Compound interest with addition formula:
[tex]A=P(1+\frac{r}{n})^{nt}+\frac{PMT(1+\frac{r}{n})^{nt}-1}{\frac{r}{n}}[/tex]where,
A = final amount
P = initial principal balance
r = interest rate
n = number of times interest applied per time period
t = number of time periods elapsed
PMT = Regular contributions (additional money added to investment)
in this example
P = 2500
r = 8% = 0.08
n = 4
t = 5 years
PMT = 2500
[tex]A=2500(1+\frac{0.08}{4})^{4\cdot5}+\frac{2500\cdot(1+\frac{0.08}{4})^{4\cdot5}-1}{\frac{0.08}{4}}[/tex]solving for A:
[tex]A=189408.29[/tex]Therefore, his investment after 5 years will be
$189,408.29
A psychologist has designed a questionnaire to measure individuals' aggressiveness. Suppose that the scores on the questionnaire are normally distributed with
a standard deviation of 90. Suppose also that exactly 10% of the scores exceed 700. Find the mean of the distribution of scores. Carry your intermediate
computations to at least four decimal places. Round your answer to at least one decimal place.
μ = 782.02 is the mean of the distribution of scores by standard deviation.
What is standard deviation in math?
A statistical measurement called standard deviation examines how far away from the mean a set of statistics is. Standard deviation, to put it simply, gauges the degree of dispersion between numbers in a data collection. The variance's square root is used to generate this metric.we have from standard normal table that
P(Z > 1.282) = 0.1
Therefore the given Z score of a score of 700 is given thus 1.282.
the z score is given:
x - μ / α = 1.282
700 - μ / 90 = 1.282
Therefore μ = (700 - 90)*1.282 = 782.02
So we have that μ = 782.02
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f(x) = log 2(x+3) and g(x) = log 2(3x + 1).(a) Solve f(x) = 4. What point is on the graph of f?(b) Solve g(x) = 4. What point is on the graph of g?(c) Solve f(x) = g(x). Do the graphs off and g intersect? If so, where?(d) Solve (f+g)(x) = 7.(e) Solve (f-g)(x) = 3.
Given
[tex]\begin{gathered} f(x)=log_2(x+3) \\ and \\ g(x)=log_2(3x+1) \end{gathered}[/tex]a)
[tex]\begin{gathered} f(x)=4 \\ \Rightarrow log_2(x+3)=4 \\ \Leftrightarrow x+3=2^4 \\ \Rightarrow x+3=16 \\ \Rightarrow x=13 \end{gathered}[/tex]The answer to part a) is x=13. The point on the graph is (13,4)
b)
[tex]\begin{gathered} g(x)=4 \\ \Rightarrow log_2(3x+1)=4 \\ \Leftrightarrow3x+1=2^4 \\ \Rightarrow3x+1=16 \\ \Rightarrow3x=15 \\ \Rightarrow x=5 \end{gathered}[/tex]The answer to part b) is x=5, and the point on the graph is (5,4).
c)
[tex]\begin{gathered} f(x)=g(x) \\ \Rightarrow log_2(x+3)=log_2(3x+1) \\ \Rightarrow\frac{ln(x+3)}{ln(2)}=\frac{ln(3x+1)}{ln(2)}] \\ \Rightarrow ln(x+3)=ln(3x+1) \\ \Rightarrow x+3=3x+1 \\ \Rightarrow2x=2 \\ \Rightarrow x=1 \\ and \\ log_2(1+3)=log_2(4)=2 \end{gathered}[/tex]The answer to part c) is x=1 and graphs intersect at (1,2).
d)
[tex]\begin{gathered} (f+g)(x)=7 \\ \Rightarrow log_2(x+3)+log_2(3x+1)=7 \\ \Rightarrow log_2((x+3)(3x+1))=7 \\ \Leftrightarrow(x+3)(3x+1)=2^7 \\ \Rightarrow3x^2+10x+3=128 \\ \Rightarrow3x^2+10x-125=0 \end{gathered}[/tex]Solving the quadratic equation using the quadratic formula,
[tex]\begin{gathered} \Rightarrow x=\frac{-10\pm\sqrt{10^2-4*3*-125}}{3*2} \\ \Rightarrow x=-\frac{25}{3},5 \end{gathered}[/tex]However, notice that if x=-25/3,
[tex]log_2(x+3)=log_2(-\frac{25}{3}+3)=log_2(-\frac{16}{3})\rightarrow\text{ not a real number}[/tex]Therefore, x=-25/3 is not a valid answer.
The answer to part d) is x=5.
e)
[tex]\begin{gathered} log_2(x+3)-log_2(3x+1)=3 \\ log_2(\frac{x+3}{3x+1})=3 \\ \Leftrightarrow\frac{x+3}{3x+1}=2^3=8 \\ \Rightarrow x+3=24x+8 \\ \Rightarrow23x=-5 \\ \Rightarrow x=-\frac{5}{23} \end{gathered}[/tex]The answer to part e) is x=-5/23
(0,1), (2,4), (4,7) (9.1)}Domain:Range:
The domain of an ordered pair are its first elements and its range are all the second elements of the ordered pair.
So, the domain ={0,2,4,9}
Range={1,4,7,1}
which of the following describes the two spheres A congruentB similarC both congruent and similarD neither congruent nor similar
The two spheres are similar since they have a proportion of their radius. This proportion is 9/6 (3/2) or 6/9 (2/3).
They are not congruent. They do not have the same radius.
Therefore, the spheres are similar.
Let f(x) = 8x^3 - 3x^2Then f(x) has a relative minimum atx=
1) To find the relative maxima of a function, we need to perform the first derivative test. It tells us whether the function has a local maximum, minimum r neither.
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3-3x^2\mright) \\ f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3\mright)-\frac{d}{dx}\mleft(3x^2\mright) \\ f^{\prime}(x)=24x^2-6x \end{gathered}[/tex]2) Let's find the points equating the first derivative to zero and solving it for x:
[tex]\begin{gathered} 24x^2-6x=0 \\ x_{}=\frac{-\left(-6\right)\pm\:6}{2\cdot\:24},\Rightarrow x_1=\frac{1}{4},x_2=0 \\ f^{\prime}(x)>0 \\ 24x^2-6x>0 \\ \frac{24x^2}{6}-\frac{6x}{6}>\frac{0}{6} \\ 4x^2-x>0 \\ x\mleft(4x-1\mright)>0 \\ x<0\quad \mathrm{or}\quad \: x>\frac{1}{4} \\ f^{\prime}(x)<0 \\ 24x^2-6x<0 \\ 4x^2-x<0 \\ x\mleft(4x-1\mright)<0 \\ 0Now, we can write out the intervals, and combine them with the domain of this function since it is a polynomial one that has no discontinuities:[tex]\mathrm{Increasing}\colon-\infty\: 3) Finally, we need to plug the x-values we've just found into the original function to get their corresponding y-values:[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(0)=8(0)^3-3(0)^2 \\ f(0)=0 \\ \mathrm{Maximum}\mleft(0,0\mright) \\ x=\frac{1}{4} \\ f(\frac{1}{4})=8\mleft(\frac{1}{4}\mright)^3-3\mleft(\frac{1}{4}\mright)^2 \\ \mathrm{Minimum}\mleft(\frac{1}{4},-\frac{1}{16}\mright) \end{gathered}[/tex]4) Finally, for the inflection points. We need to perform the 2nd derivative test:
[tex]\begin{gathered} f^{\doubleprime}(x)=\frac{d^2}{dx^2}\mleft(8x^3-3x^2\mright) \\ f\: ^{\prime\prime}\mleft(x\mright)=\frac{d}{dx}\mleft(24x^2-6x\mright) \\ f\: ^{\prime\prime}(x)=48x-6 \\ 48x-6=0 \\ 48x=6 \\ x=\frac{6}{48}=\frac{1}{8} \end{gathered}[/tex]Now, let's plug this x value into the original function to get the y-corresponding value:
[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(\frac{1}{8})=8(\frac{1}{8})^3-3(\frac{1}{8})^2 \\ f(\frac{1}{8})=-\frac{1}{32} \\ Inflection\: Point\colon(\frac{1}{8},-\frac{1}{32}) \end{gathered}[/tex]