Answer:
292.5
Step-by-step explanation:
Since they are driving 65 miles for every hour, you can multiply the 65 miles by 4.5
So:
65 * 4.5 = 292.5
Can somebody help me pls!
Answer: C
Step-by-step explanation:
Just look at a z-score table and multiply by 100.
-> (0.308538)(100) is about 30.85%
find the value of x
Answer:
See below, please
Step-by-step explanation:
[tex](2x + 9) + (4x - 3) = 90[/tex]
[tex]6x + 6 = 90[/tex]
[tex]6x = 90 - 6 = 84[/tex]
Hence
[tex]x = 14[/tex]
What is the total height of the plants that measured 1
1/8 and
1/4?
Please the answer ... Integral
Answer:
[tex]\frac{dx^{2} (x+1)S^{2} }{2(x^{2} +6x+3)^{2} }+ C[/tex]
Step-by-step explanation:
Which function has a maximum with the same maximum value as
f(x) = – |x + 3| – 2? f(x) = (x + 3)2 – 2 f(x) = –(x – 6)2 – 3
Answer:
The answer is c on edge or f(x) = 1 sqt x + 6 -2
Step-by-step explanation:
From the given two options, none of them has a function that has the same maximum value as f(x) = -|x+3|-2.
What is a function?A function is a correspondence between input numbers (x-values) and output numbers (y-values). It is used to describe an equation.
Given that:
f(x) = -|x + 3| - 2Suppose that x = c is a critical point of (x) then,
If f'(x) > 0 to the left of x = c and f'(x) < 0 to the right of x = c;
then x = c is a local maximum.If f'(x) < 0 to the left of x = c and f'(x) > 0 to the right of x = c;
then x = c is a local minimum.If f'(x) is the same sign on both sides of x = c;
then x = c and is neither a local maximum nor a local minimum.From the given equation, the critical points: x = -3
The intervals is: Increasing at -∞ < x < -3 and decreasing at -3<x<∞If we put the point x = -3 into - |x+3|-2
Then, y = -2 and it is Maximum at (-3, -2) Only f(x) = (x+3)^2 - 2 has a minimum at (-3,-2)We can therefore conclude that none of them has a function that has the same maximum value as f(x) = -|x+3|-2.
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. Gemma plans to run 5 miles her first week and increase the amount she runs each week by 20% Which of the following is closest to the total distance Genna has run after 10 weeks.
A: 115 miles. B: 130 miles
C: 138 miles. D: 145 miles
How can i prove this property to be true for all values of n, using mathematical induction.
ps: spam/wrong answers will be reported and blocked.
Proof -
So, in the first part we'll verify by taking n = 1.
[tex] \implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6} [/tex]
[tex] \implies{ \frac{1(2)(3)}{6} }[/tex]
[tex]\implies{ 1}[/tex]
Therefore, it is true for the first part.
In the second part we will assume that,
[tex] \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }[/tex]
and we will prove that,
[tex]\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}[/tex]
[tex] \: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}[/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6} [/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6} [/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}[/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6} [/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6} [/tex]
[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6} [/tex]
Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.
_______________________________
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Eva filled a bucket with 7 gallons of water. A few minutes later, she realizes only 1 3/5 of water remained. How much water had leaked out of the bucket?
Answer:
[tex]6 \frac{2}{5}[/tex]Step-by-step explanation:
[tex]7 - 1 \frac{3}{5} = 6 \frac{2}{5}[/tex]
i need help
Simplify the expression 63 + 5(4 − 2).
28
36
226
234
Answer:
226
Step-by-step explanation:
Given:
Simplify 6^3+5(4-2)
Note:
I think you meant 6^3 because if you solve 63+5(4-2):
63+5(4-2)
63+5 * 2
63 + 10
73
Solve:
6^3 + 5(4 - 2 )
6^3 + 5 x 2
6 x 6 x 6 = 216
226 + 5 x 2
5 x 2 = 10
216 + 10 = 226
~Lenvy~
Determine the values of k for which the function f(x) = 4x^2-3x + 2kx + 1 has two zeros. Check these values in the original equation.
k must be greater than or equal to 22.75 to have two different zeros.
How to determine the value of missing coefficient in second order polynomials
Second order polynomials are algebraic expressions that observe the following form:
[tex]p(x) = a\cdot x^2 + b\cdot x + c[/tex] (1)
Where:
a, b, c - Coefficientsx - Independent variableFor polynomials of the form p(x) = 0, we can infer the nature of their roots by applying the following discriminant:
d = b² - 4 · a · c (2)
According to (2), there are three cases:
If d < 0, then there are two conjugated complex roots.If d = 0, then the two roots are the same real number.If d > 0, then the two roots are two distinct real numbers.Now we have the following discriminant case:
-(3 + 2 · k)² - 4 · (1) · (4) ≠ 0
-(9 + 6 · k + 4 · k²) - 16 ≠ 0
-9 - 6 · k - 4 · k² - 16 ≠ 0
4 · k²+ 6 · k +25 ≠ 0
This characteristic polynomial has two conjugated complex roots, then we conclude that all values of k must positive or negative, but never zero. By graphng tools we find that k must be greater than or equal to 22.75 to have two different zeros.
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Find the missing information for the triangle.
*not drawn to scale
• Make sure to find the missing angle measure and the 2 missing side
lengths.
missing angle:
180° - 90° - 30°
180° - 120°
60°
missing sides:
(a)
[tex]\rightarrow \sf tan(x)= \dfrac{opposite}{adjacent}[/tex]
[tex]\rightarrow \sf tan(30)= \dfrac{4}{adjacent}[/tex]
[tex]\rightarrow \sf adjacent= \dfrac{4}{tan(30)}[/tex]
[tex]\rightarrow \sf adjacent= 4\sqrt{3}[/tex]
[tex]\rightarrow \sf adjacent= 6.93 \ cm[/tex]
(b)
[tex]\sf \rightarrow sin(x)= \dfrac{opposite}{hypotensue}[/tex]
[tex]\sf \rightarrow sin(30)= \dfrac{4}{hypotensue}[/tex]
[tex]\sf \rightarrow hypotensue= \dfrac{4}{ sin(30)}[/tex]
[tex]\sf \rightarrow hypotensue= 8 \ cm[/tex]
Answer:
m∠X = 60°
BX = 8 cm
BM = 4√3 cm
Step-by-step explanation:
The sum of the interior angles of a triangle is 180°
Given:
m∠B = 30°m∠M = 90°⇒ m∠B + m∠M + m∠X = 180°
⇒ 30° + 90° + m∠X = 180°
⇒ 120° + m∠X = 180°
⇒ m∠X = 180° - 120°
⇒ m∠X = 60°
Using the sine rule to find the side lengths:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
(where A, B and C are the angles, and a, b and c are the sides opposites the angles)
Given:
m∠X = 60°m∠B = 30°m∠M = 90°MX = 4 cm[tex]\implies \dfrac{4}{\sin 30\textdegree}=\dfrac{BX}{\sin 90\textdegree}=\dfrac{BM}{\sin 60\textdegree}[/tex]
[tex]\implies BX=\sin 90\textdegree \cdot\dfrac{4}{\sin 30\textdegree}[/tex]
[tex]=1 \cdot \dfrac{4}{\frac12}[/tex]
[tex]=1 \cdot 4 \cdot 2[/tex]
[tex]=8 \textsf{ cm}[/tex]
[tex]\implies BM=\sin 60\textdegree \cdot\dfrac{4}{\sin 30\textdegree}[/tex]
[tex]=\dfrac{\sqrt{3}}{2}\cdot \dfrac{4}{\frac12}[/tex]
[tex]=\dfrac{\sqrt{3}}{2}\cdot 4 \cdot 2[/tex]
[tex]=4\sqrt{3} \textsf{ cm}[/tex]
Find the area if the pentagon. I’ll mark the brainiest :)
Answer:
688.19 inches
Step-by-step explanation:
will give brainliest but it Has to be correct.
Measure the thumbtack to the nearest inch.
6/8 inches
7/8 inches
3/8 inches
5/8 inches
Answer: 6/8 inch
Step-by-step explanation:
Solve for x please :)
Answer:
see explanation
Step-by-step explanation:
look photo
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
The given pair of angles form linear pair, therefore their sum is equal to 180°
that is :
[tex]\qquad \sf \dashrightarrow \:2(x - 26) + 3x + 2 = 180[/tex]
[tex]\qquad \sf \dashrightarrow \:2x - 52 + 3x + 2 = 180[/tex]
[tex]\qquad \sf \dashrightarrow \:5x - 50 = 180[/tex]
[tex]\qquad \sf \dashrightarrow \:5x = 230[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 46 \degree[/tex]
Have a great day ~
According to the line plot how many apples weigh 5/8 of a pound
Answer:
Answer:4 apples weigh 5/8 pound.
Step-by-step explanation:
Answer:
2(−5) − 10 = 2(0)
Step-by-step explanation:
If you substitute the values x = 0 and y = −5 into the second equation, you get a false statement
How can you tell that (496 + 77 + 189) x 10 is twice as large as (496 + 77 +189) x 5 without doing complicated calculations?
Answer:
Because 10 is twice as large as 5.
Step-by-step explanation:
You randomly draw twice from this deck of cards
0 с G|F. D C G
What is the probability of not drawing a C, then not drawing a C,
without replacing the first card? Write your answer as a decimal
rounded to the nearest hundredth.
The probability of not drawing C in neither draw is P = 0.5
How to get the probability?
All the cards have the same probability of being drawn, in this case, our set of cards is {F, D, C, G}
The probability of not drawing C is equal to the probability of drawing F, D or G. So we have 3 options out of 4, then the probability is:
p = 3/4.
Now we draw another, this time there are 3 cards, one of these is C, and the other two cards are not C. Then the probability of not drawing C again is equal to 2 over 3.
q = 2/3.
The joint probability (for both of these events to happen) is equal to the product of the individual probabilities:
P = p*q = (3/4)*(2/3) = 0.5
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Sita saves Rs. 1 today, Rs. 2 the next day, Rs. 4 the succeeding day and so on (each saving being twice of the preceding one). What will be total saving in two weeks time?
a
Answer:
Rs. 32767
Step-by-step explanation:
Because the amount is doubling every day, we can use the expression 1*2^15-1 because there is 1 to start with. Also cool trick! if you need to do 2^1+2^2+2^3+....+2^x, it will be equal to 2^(x+1)-1. So:
2^15-1
32768-1
32767
The loudness (L) of sound in decibels is related to intensity (I)measured in watts per square centimeter by the equation: L = 10log( I 10-16 ). Find the loudness of a whisper at 10-12 W/cm2. A) 35 decibels B) 40 decibels C) 45 decibels D) 50 decibels
The function L= 10 log(I/10^-16) is a logarithmic equation
The loudness of the whisper is 40 decibels
How to determine the loudness?The function of the loudness is given as:
L= 10 log(I/10^-16)
When the intensity is 10^-12, the equation becomes
L= 10 log(10^-12/10^-16)
Evaluate the quotient
L= 10 log(10^4)
Apply the rule of logarithm
L= 10 * 4
Evaluate the product
L = 40
Hence, the loudness of the whisper is 40 decibels
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The square root of 7^16 is equal to 7^n for some positive integer n. Find n.
[tex]\sqrt{7^{16}} = 7^n\\\\\implies \left(7^{16}\right)^{\tfrac 12} = 7^n\\\\\implies 7^{\left(\tfrac 12 \times 16\right)}=7^n\\\\\implies 7^8 = 7^n\\\\\implies \ln 7^8 = \ln 7^n\\\\\implies 8\ln 7 = n \ln 7\\\\\implies n =8[/tex]
An airplane flies with a constant speed
of 840 km/h. How far can it travel in
1 hour?
Answer:
840 km
Step-by-step explanation:
The speed expression ...
840 kilometers per hour
means the plane files 840 kilometers in each hour.
In 1 hour, it will travel 840 km.
Vocabulary
1. Volume: A measure of ________ occupied by a __________-________________ figure.
1. Base: The __________ on which an object _______.
1. Height: The ______ distance from top to bottom, creates a ___-degree angle with the base.
1. Inverse Operation: The ________ of a math operation; the opposite of addition is ________ and the opposite of multiplication is ________.
1. Diameter: A ________ line going from one side of a ______ to the other through the _______.
1. Radius: The distance from the ______ to the ______ of a ______; _____ of the diameter.
Volume of a Cylinder
A ____________ is a _____________________ object with a _________________ base and top.
To find the ____________ of a ______________ we use the following formula:
Answer:
Step-by-step explanation:
. Volume: A measure of _space occupied by a _three dimensional _ figure.
1. Base: The surface on which an object stands on.
1. Height: The _vertical distance from top to bottom, creates a _90° degree angle with the base.
1. Inverse Operation: The opposite of a math operation; the opposite of addition is subtraction and the opposite of multiplication is division.
1. Diameter: A straight line going from one side of a point on a circle to the other through the _center.
1. Radius: The distance from the center to the point of a circle;or half of the diameter.
Volume of a Cylinder
A cylinder is a three dimensional object with a circular base and top.
To find the volume of a cylinder we use the following formula:πr²h
What is the approximate volume of a cone with a height of 9 ft and radius of 3 ft? Use 3.14 to approximate pi, and express your final answer to the nearest hundredth Enter your answer as a decimal in the box. ft3
How do you know that the Pythagorean Theorem is true?
The fact that the angles in a triangle add up to 180 indicates that it is actually a square). There are also four right triangles, each with a base and a height. The Pythagorean Theorem is reached when a2 + b2 = c2.
A perfect score on a test with 25 questions is 100. Each question is worth the same number of points. How many points is each question on the test worth
Answer:
4
Step-by-step explanation:
100 divided by 25 equals 4.
Need help on number 10
If tan C is 3/4, find the sin C.
Answer:
sin C = 3/5
Step-by-step explanation:
see image.
It helps to draw a picture. Tan C is the ratio of the OPP/ADJ.
Pythagorean theorem or if you know Pythagorean triples are a shortcut to find the hypotenuse.
Once you know the hypotenuse, use the ratio for sine to solve the question. Sine is OPP/HYP.
see image.
spherical water tank of radius R = 5m is emptied through a small circular hole of radius r = 0.03 m at the bottom. The top of the tank is open to the atmosphere. The instantaneous water level h in the tank (measured from the bottom of the tank, at the drain) can be determined from the solution of the following ODE:
dh /dt =r²(2gh)^0.5/ 2hR-h²
where g = 9.81 m/s². If the initial (t = 0) water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.
Answer:
water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.
Step-by-step explanation:
water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.
[tex]\large \rm \sum \limits_{n = 0}^ \infty \frac{( { - 1)}^{1 + 2 + 3 + \dots + n} }{(2n + 1 {)}^{2} }[/tex]
The sum we want is
[tex]\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots[/tex]
where [tex]T_n=\frac{n(n+1)}2[/tex] is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as
[tex]\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)[/tex]
For convenience, I'll use the abbreviations
[tex]S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}[/tex]
[tex]{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}[/tex]
for m ∈ {1, 2, 3, …, 7}, as well as the well-known series
[tex]\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}[/tex]
We want to find [tex]S_1-S_3-S_5+S_7[/tex].
Consider the periodic function [tex]f(x) = \left(x-\frac12\right)^2[/tex] on the interval [0, 1], which has the Fourier expansion
[tex]f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}[/tex]
That is, since f(x) is even,
[tex]f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)[/tex]
where
[tex]a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}[/tex]
[tex]a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}[/tex]
(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)
Expand the Fourier series to get sums resembling the [tex]S'[/tex]-s :
[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)[/tex]
which reduces to the identity
[tex]\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'[/tex]
Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution
[tex]\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}[/tex]
It turns out that
[tex]{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7[/tex]
so we're done, and the sum's value is [tex]\boxed{\dfrac{\pi^2}{8\sqrt2}}[/tex].
Find the area of sector RST Enter your answer in terms of a fraction of it and rounded to the nearest
hundredth.
Fort nite battle pass is 8 dollars
Question 2:
If the following frequency distribution shows the average number of students per teacher in the 50 major cities of Pakistan
Class Limits Frequency
9-11 3
12 – 14 5
15 – 17 12
18 – 20 18
21 – 23 8
24 – 26 4
Table 1
Determine
• Range
• Mean
• Median
• Mode
• Standard Deviation
• Relative Dispersion
• Variance
• Kurtosis
With the frequecy distribution shown in the 50 cities of pakistan,
range = 18mean = 18.1median = 19.8333mode = 19.125kurtosis = 2.7508Standard deviation = 3.75How to find the Range= highest value - lowest value
= 26.5 - 8.5
= 18
How to find the mean= ∑ f x / ∑ f
= ∑ f x / N
= 905 / 50
= 18.1
median
= lower limit + ( N/2 - C ) * h / ( frequency of the class interval )
C = cumulative frequency preceeding to the median class frequency
h = class interval
= 18.5 + ( 50 / 2 - ( 5 + 12 ) ) * 3 / 18
= 18.5 + 1.3333
= 19.8333
How to find the modeThe mode is the value with the highest frequency occurence. This is under class 18 - 20
mode = lower limit + ( ( f1 - f0 ) / (2*f1 - f0 - f2 ) ) * h
f1 = fequency of the modal class
f0 = freqency of the preceeding modal class
f2 = frequency of the next modal class
h = class interval
= 18.5 + ( ( 18 - 12 ) / (2 * 18 - 12 - 8 ) ) * 3
= 18 + ( 0.375 ) * 3
= 19.125
How to find the standard deviation= sqrt ( 1 / N ∑ f ( x - x' )^2 )
= sqrt (1 / 50 * 706.5
= 3.7589
How to solve for relative dispersion= standard deviation / mean
= 3.7589 / 3
= 1.2530
What is the variance?= ( standard deviation )^2
= ( 3.7589 )^2
= 14.1293
How to solve for kurtosis= ∑ f ( x - x' )^4 / ( N * ( standard deviation )^4 )
= 27459.405 / ( 50 * 3.7589^4 )
= 2.7509
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