The process of Graphing a exponential decay function is shown in detail below.
There are two types of exponential equations: exponential decay and exponential growth. The fundamental shape of an exponential function and its corresponding graphs must be understood in order to fully comprehend the distinctions between exponential growth and decay. The fundamental distinction between the two is that in an exponential decay connection, the output values rise quickly as the input value rises, but in an exponential growth relationship, the output rises noticeably faster as the input value rises. Furthermore, neither of the two functions is linear, hence their rate of change is not constant. What distinguishes one equation from the other when they both have the same precise form, y = abx?
The value of the constant "b" is the fundamental factor in deciding whether the exponential function is one of growth or decay:
If function y = ab^x and b > 1then the function is an exponential growth function.If the function y = ab^x and 0<b<1 then the function is an exponential decay function. As noted before, "b" can never be exactly 1.In this way we can graph a exponential decay function
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Which of the following best describes the set of complex numbers?
OA. The set of all numbers of the form a+bi, where a and bare any
real numbers and i equals -1
B. The set of all numbers of the form abi, where a and bare any real
numbers and i equals 1
C. The set of all numbers of the form a+bi, where a and b are any
real numbers and i equals √-1
OD. The set of all numbers of the form abi, where a and b are any real
numbers and / equals -1
Answer:
C
Step-by-step explanation:
The correct statement is option C.
What is complex number?A real number and an imaginary number are effectively combined to create a complex number. The complex number is written as a+ib, where a and ib are real and imaginary numbers, respectively. Additionally, i = √-1 and both a and b are real numbers.
Since we know that
Complex number is of the form a+ib
Where,
a is real number belongs to real axis
And b is also a real number belongs to imaginary axis.
And the value of i = √-1
Thus,
The set of all numbers of the form a+bi, where a and b are any
real numbers and i equals √-1 is the correct statement.
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5
The recursive formula for a geometric sequence is given below. What is the third term in the sequence?
f(1) = 2
f(n) = 3f(n − 1)
-
OA. 18
OB. 54
OC. 162
OD. 12
Reset
Submit
[tex]\Large{ \boxed{ \tt{ \blue{A}}}}\Large{ \boxed{ \tt{ \pink{N}}}}\Large{ \boxed{ \tt{ \green{S}}}}\Large{ \boxed{ \tt{ \purple{W}}}}\Large{ \boxed{ \tt{ \red{E}}}}\Large{ \boxed{ \tt{ \pink{R}}}}[/tex]
[tex] \: \: [/tex]
A. 18[tex] \: \: [/tex]
[tex]\Large{ \boxed{ \tt{ \blue{S}}}}\Large{ \boxed{ \tt{ \pink{O}}}}\Large{ \boxed{ \tt{ \green{L}}}}\Large{ \boxed{ \tt{ \purple{U}}}}\Large{ \boxed{ \tt{ \red{T}}}}\Large{ \boxed{ \tt{ \pink{I}}}}\Large{ \boxed{ \tt{ \blue{O}}}}\Large{ \boxed{ \tt{ \pink{N}}}}[/tex]
[tex] \: \: [/tex]
[tex] \mathtt{f(1) = 2}[/tex][tex] \tt{f(2) \: = \: 3f \: (1) \: = \: 3(2) \: = \: 6}[/tex][tex] \tt{f(3) \: = \: 3f(2) \: = \: 3(6) \: = \: 18}[/tex][tex]\color{pink}─────────────────────────────────────[/tex]
keep learning
Refer to the figure to complete the following item.
Given:
If m = 60° and m = 30°, then 3 =
45
15
30
If m VB = 60° and m BS = 30°, then m ∠3 =15°.
What is tangent?A tangent is a line passing by touching the perimeter of the circle, perpendicular to the line joining the center and touch point.
Given, PB tangents PV, PU secants
If m VB = 60° and m BS = 30°, then m ∠3 =
The measurement of the external angles is the semi difference of the arcs it consists.
∠3 = 1/2 (60 - 30)
∠3 = 15°
Thus, If m VB = 60° and m BS = 30°, then m ∠3 =15°.
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Match the scatter plot with its description
weak and positive
moderate and negative
strong and negative
strong and positive
weak and negative
moderate and positive
A line of best fit was drawn for 16 data points. What is the maximum number
of these data points that may not actually be on the line?
OA. 14
OB. 13
O C. 15
OD. 16
The maximum number of data points that may not actually be on the line is 16. so, the correct option is D.
How to find the line of best fit?We know that a line of best fit is basically a straight line drawn for a given data that may or may not pass through data points.
It is given that the line is drawn for 16 data points.
Hence, the line may or may not pass through all these 16 points.
Therefore, the maximum number of data points that may not actually be on the line is 16.
so, the correct option is D.
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According to American Airlines, flight 71098 from New York to Los Angeles is on time 88.9% of the time. Assume that we randomly select 150 flights, use the normal approximation to the binomial to do the following:
a) approximately the probability that exactly 124 flights are on time.
b) approximate the probability that between 113 and 130 flights ,inclusive, are on time.
Using the normal approximation to the binomial, it is found that the probabilities are given as follows:
a) 0.0055 = 0.55%.
b) 0.2296 = 22.96%.
Normal Probability DistributionThe z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score measures how many standard deviations the measure is above or below the mean. Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].The parameters of the binomial distribution are given as follows:
n = 150, p = 0.889.
Hence the mean and the standard deviation of the approximation are:
[tex]\mu = E(X) = np = 150 x 0.889 = 133.35[/tex].[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150(0.889)(0.111)} = 3.8473[/tex]Item a:
Using continuity correction, the probability is P(123.5 < X < 124.5), which is the p-value of Z when X = 124.5 subtracted by the p-value of Z when X = 123.5, hence:
X = 124.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{124.5 - 133.35}{3.8473}[/tex]
Z = -2.3
Z = -2.3 has a p-value of 0.0107.
X = 123.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{123.5 - 133.35}{3.8473}[/tex]
Z = -2.56
Z = -2.56 has a p-value of 0.0052.
Hence the probability is 0.0107 - 0.0052 = 0.0055 = 0.55%.
Item b:
The probability is P(112.5 < X < 130.5), which is the p-value of Z when X = 130.5 subtracted by the p-value of Z when X = 112.5, hence:
X = 130.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130.5 - 133.35}{3.8473}[/tex]
Z = -0.74
Z = -0.74 has a p-value of 0.2296.
X = 112.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{112.5 - 133.35}{3.8473}[/tex]
Z = -5.42
Z = -5.42 has a p-value of 0.
Hence the probability is 0.2296 - 0 = 0.2296 = 22.96%.
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Managerial Accounting
Answer:
1400 units
$26,600
1700 units
$32,300
Step-by-step explanation:
yw :))
PLEASEEEE PLEASEEEEEEEEE HELPPPPPPPPPP
how do i solve this equation?
Answer:
360
Step-by-step explanation:
using the definition
n [tex]P_{r}[/tex] = [tex]\frac{n!}{(n-r)!}[/tex]
where n! = n(n - 1)(n - 2)..... × 3 × 2 × 1
then
6[tex]P_{4}[/tex]
= [tex]\frac{6!}{(6-4)!}[/tex]
= [tex]\frac{6!}{2!}[/tex]
= [tex]\frac{6(5)(4)(3(2)(1)}{2(1)}[/tex] ← cancel 2(1) on numerator / denominator
= 6 × 5 × 4 × 3
= 360
can someone find the value of x?
Answer:
[tex]\boxed{\sf x=12}[/tex]
Step-by-step explanation:
By applying Similar Triangles Theorem:-
[tex]\sf \cfrac{x}{6} =\cfrac{6}{3}[/tex]
We'll multiply both sides of the equation by 6, the LCM of 6,3.
[tex]\sf x=2 \times 6[/tex]
[tex]\sf x=12[/tex]
What is the range of the function g(x) = –3sec(2x + 4) – 1?
(–∞, –2] ∪ [0, ∞)
(–∞, –4] ∪ [0, ∞)
(–∞, –4] ∪ [2, ∞)
(–∞, –5] ∪ [1, ∞)
The range of the function g(x) will be (–∞, –4] ∪ [2, ∞). Option C is correct.
What is the difference between domain and range?The domain denotes all potential x values, while the range denotes all possible y values.
When we plot the graph of the given function we will get the maximum and the minimum value till we get the function plot. The difference in the value of those coordinates is the range of the function.
The range of the function g(x) will be (–∞, –4] ∪ [2, ∞).
Hence option C is correct.
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❗️URGENT❗️
Chase has won 80% of the 30 football video games he has played with his brother, Jacob. If Chase and Jacob continue to play the video game, how many more games in a row goes Chase have to win to achieve an 85% win percentage?
Answer:
10
Step-by-step explanation:
24/30 = 0.8
24+x/30+x =0.85
10=x
If you want more clarification just ask
Total games Chase has to win in a row to achieve an 85% win percentage would be 10.
We can write the total number game won by Chase as -
w = 80% of 30
w = 80/100 x 30
w = 4/5 x 30
w = 24
Assume that he wins {x} games after winning 24 games to achieve 85% win percentage. Now, we can write that out of (30 + x) number of games played, Chase won (24 + x) games to achieve 85% win percentage. So, we can write that -
(24 + x)/(30 + x) = 85/100
(24 + x)/(30 + x) = 0.85
(24 + x) = 0.85(30 + x)
24 + x = 25.5 + 0.85x
0.15x = 1.5
x = 10
Total games Chase has to win in a row to achieve an 85% win percentage would be 10.
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Help with this math problem
Answer:
The minimum cost of the X-ray machines is 12,197dollars.
Step-by-step explanation:
First, we check whether it’s a Quadratic Equation or not
The general form of the Quadratic equation [tex]f(x) =[/tex] [tex]ax^{2} + bx + c[/tex]
Let’s compare it with the given equation, we get
a = 1
b = -520
c = 79,79
Hence, it’s a quadratic equation.
Now, we will use the VERTEX FORMULA as we are asked to find the ‘minimum’ unit cost.
X = [tex]\frac{-b}\[2a[/tex] = [tex]\frac{-(-520)}\[2(1)[/tex] = [tex]260[/tex]
So , number of X-ray machines are = 260 (value of x)
To find the minimum unit cost, plug the value of x into the given equation, and we get
[tex]f(260) = (260)^{2} -520(260) + 79,797[/tex]
[tex]=12,197[/tex]
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Which absolute value functions will be narrower than the parent function, f(x) = |x|? Check all that apply.
f(x) = |x|
f(x) = |x – 2|
f(x) = |x| + 3
f(x) = 2.9|x|
f(x) = 1.2|x + 8|
f(x) = 0.7|x| – 3.2
The correct answer is option 4 which is f(x) = 2.9|x| is narrower than the parent function f(x) = |x|
What is a function?A function is defined as the expression that set up the relationship between the dependent variable and independent variable.
Given parent function is f(x)=|x|
Now we have to find which function from given choices will be narrower than the parent function.
Notice that adding or subtracting some number from the parent function only shifts the graph up, down, and left of the right side.
But that will not make the function narrower or broader.
So f(x) = |x – 2| and f(x) = |x| + 3, can't be the answer.
Multiplying by some positive real number which is more than 1, makes the function narrower.
Only f(x) = 2.9|x| from remaining choices fits that case.
Hence correct answer is option 4 which is f(x) = 2.9|x| is narrower than the parent function f(x) = |x|
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Answer:
D,E
Step-by-step explanation:
In 2010, the amount in Raja's account was 40 Lakhs. He debited 8000 in 2011, and in 2012, he debited 2,07,000. In 2013, he credited 1,16,000, and in 2014 he credited 12,000. How much more he has to credit to be worth what it was at at the start of 2010
How much more he has to credit to be worth what it was at at the start of 2010 is: 3,657,000.
Additional credit amountFirsts step
Total amount debited=8,000+207,000
Total amount debited=215,000
Total amount credited=116,000+12,000
Total amount credited=128,000
Second step
Additional amount to credit:
40 Lakhs is 4,000,000
Hence:
Additional amount to credit=4,000,000- (215,000+128,000)
Additional amount to credit=4,000,000-343,000
Additional amount to credit=
Therefore how much more he has to credit to be worth what it was at at the start of 2010 is: 3,657,000.
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Given any two events, E, and E₂, what does the probability P(E, U E₂) represent?
Answer:
P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2)
Step-by-step explanation:
As we know that, if A and B are two events then
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⇒ P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2)
Seismology In 1812, an earthquake of magnitude 7.9 shook New Madrid,
Missouri. Compare the amount of energy released by that earthquake to the
amount of energy released by each earthquake below.
magnitude 9.5 in Valdivia, Chile, in 1960
which equation has the solution x=6
Answer:
need more info
Step-by-step explanation:
what I can tell you is that when you change all the x's in the equation to 6s, both sides of the equation will equal eachother.
quest
The chart shows how many people have signed up to go on a field trip each day. 62 students are allowed to go on the field trip. On
which day would you expect that number to be reached?
D)
10
Days People
1
26
2
30
3
34
4
38
5
42
6 46
og php?totalQuestions-10&testid-5
strand-5716&element-18789&condition-random #
Answer:
Day 5 ................
The power 3 Superscript negative 3 equals StartFraction 1 Over 27 EndFraction . Which expression is equivalent to 3 Superscript negative 3?
Applying the negative exponent, the equivalent expression [tex]3^{-3}[/tex] is 1/27.
What are exponents?The exponents of a number are defined as the representation of a number that shows how many times a number is multiplied by itself.
When we have a negative exponent, we use a fraction, with the term with the exponent going to the denominator.
Hence, the equivalent expression is:
[tex]3^{-3} = \dfrac{1}{3^3} = \dfrac{1}{27}[/tex]
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Somebody please assist me here
The base case of [tex]n=1[/tex] is trivially true, since
[tex]\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)[/tex]
but I think the case of [tex]n=2[/tex] may be a bit more convincing in this role. We have by the inclusion/exclusion principle
[tex]\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)[/tex]
with equality if [tex]E_1\cap E_2=\emptyset[/tex].
Now assume the case of [tex]n=k[/tex] is true, that
[tex]\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)[/tex]
We want to use this to prove the claim for [tex]n=k+1[/tex], that
[tex]\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]
The I/EP tells us
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)[/tex]
and by the same argument as in the [tex]n=2[/tex] case, this leads to
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})[/tex]
By the induction hypothesis, we have an upper bound for the probability of the union of the [tex]E_1[/tex] through [tex]E_k[/tex]. The result follows.
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]
the segments ab and cd are graphed on a coordinate plane. The endpoints of ab are at (9,4) and (9,20). CD is parallel to the x-axis, and one of its endpoints is at (5,8). if ab=cd and cd is entirely in the first quadrant, what is the other endpoint of cd
Answer:
21, 8
Step-by-step explanation:
The second endpoint is of the form (X, 8) since it's parallel to the x axis.
Given the congruence of the two segments, we can tell that its length is 16 units (since [tex]|4-20| = |-16|= 16[/tex])
At this point the two possible endpoints for the segment are [tex](5-16; 8)[/tex] or [tex](5+16, 8)[/tex]. The fact that the segment has to sit in the first quadrant rules out the first option (it's endpoint will be at (-11, 8) which will set most of it in the second quadrant) and we're left with (21, 8)
If f(x)=x+2 and g(x)=x2+1 find:a. f(g(x)) b. g(f(x))
Answer:
Step-by-step explanation:
Answer:
x² + 3 and x² + 4x + 5
Step-by-step explanation:
(a)
to find f(g(x)) substitute x = g(x) into f(x)
f(g(x))
= f(x² + 1)
= x² + 1 + 2
= x² + 3
(b)
to find g(f(x)) substitute x = f(x) into g(x)
g(f(x))
= g(x + 2)
= (x + 2)² + 1 ← expand factor using FOIL
= x² + 4x + 4 + 1
= x² + 4x + 5
Solve the equation.
6(x−1)6/7=12
Answer:
Linear Equations In One Variable =
[6(x-1)6] / 7 = 12
[(6x - 6)6] = 84
[36x - 36] = 84
36x = 84 + 36
36x = 120
x = 120/36
x = 10/3
equation solved (Answer : 10/3)
Need number 2 please!!
Answer:
A
Step-by-step explanation:
note that i = [tex]\sqrt{-1}[/tex]
given
- 1 + 2i[tex]\sqrt{3}[/tex]
= - 1 + [tex]\sqrt{2^2(-1)3}[/tex]
= - 1 + [tex]\sqrt{-12}[/tex]
Please help quick!!
Answer:
the answer will be similar, because if you divide 2, 3, 2.25 by 3 you will get the answer.
Hi can someone please help me with this geometry problem? Thanks!
A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry.
What is a triangle?A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°.
The triangles ΔEST and ΔEFD are similar triangles, therefore, we can write,
[tex]\dfrac{ES}{EF} = \dfrac{ET}{ED} = \dfrac{ST}{FD}[/tex]
Since S and T are midpoints of EF and ED, the lines will be divided into two equal parts. Therefore,
[tex]\dfrac{ES}{EF} = \dfrac{ET}{ED} = \dfrac{ST}{FD}= \dfrac12[/tex]
Therefore, we can write it as,
[tex]FD = 2 (ST)[/tex]
In ΔEST and ΔTDR
∠T ≅ ∠T {Vertical angles}
ET ≅ TD {T is the midpoint of ED}
∠SET ≅ ∠TDR {Alternate interior angles}
Therefore, ΔEST ≅ ΔTDR.
Since the two triangles are equal we can write,
ST ≅ TR
Further, it can be written as,
FD = 2(ST)
FD = ST + ST
FD = ST + TR
FD = SR
Hence, FD≅SR.
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three-quarters of a pile of bricks were used for a certain project. when two thirds of the reminder had been used, 50 bricks were left. how many bricks were there in the original pile?
Answer:
sorry I don't know the answer
what is the length of the squares side?
Answer:
side ≈ 4.24= [tex]\frac{6}{\sqrt{2} }[/tex]
Step-by-step explanation:
sides a = b
[tex]6^{2} =a^{2} +a^{2}[/tex]
[tex]6^{2} =2a^{2}[/tex]
[tex]a^{2}= \frac{6^{2} }{2} =\frac{36}{2} =18[/tex]
[tex]a=\sqrt{18} =4.24=\sqrt{(9)(2)} =\frac{6}{\sqrt{2} }[/tex]
Hope this helps
Simplify:
(x-1)+(12–7.5x)
Answer:
[tex]-6.5x+11[/tex]
Step-by-step explanation:
Expand The Brackets:
[tex]x-1+12-7.5x[/tex]
[tex]=x+(-1)+12+(-7.5x)[/tex]
Combine Like Terms:
[tex]=x+(-1)+12+(-7.5x)[/tex]
[tex]=(x+-7.5x)+(-1+12)[/tex]
Answer:
[tex]-6.5x+11[/tex]
I Hope This Helps
easy:
Clara buy: 300 apples, 74 potato, 15 eggs e 2 phone.
how many things did he buy in all?
Hard:
There are 92 boys in a school, the girl 100.How many more girl are there than boys?
Answer:
easy = 391 items
hard = there are 8 more girls than there are boys
