The sum of the 24 terms of the arithmetic series is 900
How to solve:
To find the sum of an arithmetic series, we can use the formular:
Sn = (n/2)(2a + (n-1)d)
Where;
Sn is the series sum, n is the number of terms, a is the first term, and d is the common difference.
The first term (a) in this scenario is 3, and the common difference (d) is 3. With n equal to 24, we want to find the sum of the first 24 terms.
When the values are entered into the formula, we get:
The sum of the first 24 terms of the arithmetic series is 900 because;
S24 = (24/2)(2(3) + (24-1)(3))
= 12(6 + 23(3))
= 12(6 + 69)
= 12(75)
= 900.
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Evaluate functions from their graph g(−9)=
See the Correct Answer Attached!
Evaluating the function from the given graph, g(-9) = 1.
How to Evaluate a Function?To find the value of a function for a given x-value using a given graph, trace the x-value against the corresponding y-value on the y-axis.
To evaluate the function from the graph given, for g(-9), find the y-value that corresponds to the x-value of -9.
The graph shows that when x = -9, the corresponding y-value is 1.
Therefore, g(-9) = 1.
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solve this pleas i need urgent answer
Answer:
f(n)={n/2 = if n is even= n={2/2
{3n+1 {3(2)+1 =7
Step-by-step explanation:
if n is even u can put any even no in that place and u can get odd no or even no.(i think)
If A is a 2 × 2 matrix, then A × I =
and I × A =
Since the multiplication between two matrices is not commutative, then [tex]\vec A\, \times\,\vec I \ne \vec I \,\times \,\vec A[/tex], regardless of the dimensions of [tex]\vec A[/tex].
Is the product of two matrices commutative?
In linear algebra, we define the product of two matrices as follows:
[tex]\vec C = \vec A \,\times \vec B[/tex], where [tex]\vec A \in \mathbb{R}_{m\times p}[/tex], [tex]\vec B \in \mathbb{R}_{p\times n}[/tex] and [tex]\vec C \in \mathbb{R}_{m \times n}[/tex] (1)
Where each element of the matrix is equal to the following dot product:
[tex]c_{ij} = \left[\begin{array}{cccc}a_{i1}&a_{i2}&\ldots&a_{ip}\end{array}\right]\,\bullet\,\left[\begin{array}{ccc}b_{1j}\\b_{2j}\\\vdots\\b_{pj}\end{array}\right][/tex], where 1 ≤ i ≤ m and 1 ≤ j ≤ n. (2)
Because of (2), we can infer that the product of two matrices, no matter what dimensions each matrix may have, is not commutative because of the nature and characteristics of the definition itself, which implies operating on a row of the former matrix and a column of the latter matrix.
Such "arbitrariness" means that resulting value for [tex]c_{ij}[/tex] will be different if the order between [tex]\vec A[/tex] and [tex]\vec B[/tex] is changed and even the dimensions of [tex]\vec C[/tex] may be different. Therefore, the proposition is false.
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solve 4/7- (-4/7)
A. 8/7
B.-8/7
C.-4/7
D.4/7
Answer:
A. 8/7
Step-by-step explanation:
Step 14/7 - (-4/7) --> 4/7 + 4/7
The two minutes turn into a plus, removing the bracket.
Step 24/7 + 4/7 = 8/7
Add the two fractions together.
Answer:
A. 8/7
Step-by-step explanation:
What is the area of an equilateral triangle having side 'a' units?
Answer:
[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]
Step-by-step explanation:
Since an equilateral has all of the sides equal, we can find the height of triangle using: [tex]a^2+b^2=c^2[/tex]. I attached a diagram which should explain how I got the dimensions of the three sides. Using the information from the diagram we get the equation:
[tex]h^2+(\frac{a}{2})^2=a^2[/tex]
Subtract a^2 from both sides
[tex]h^2=a^2-(\frac{a}{2})^2[/tex]
Take the square root of both sides
[tex]h = \sqrt{a^2-(\frac{a}{2})^2}[/tex]
If you know the area of a triangle, it's: [tex]\frac{1}{2}bh[/tex]. In this case the base=a, and the height is what we defined above. Using this we get:
[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a}{2})^2}[/tex]
We can distribute the exponent over the division to get:
[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a^2}{4})[/tex]
Now we can rewrite a^2 as 4a^2/4
[tex]A = \frac{a}{2}*\sqrt{\frac{4a^2}{4}-(\frac{a^2}{4})[/tex]
Now add the two fractions:
[tex]A = \frac{a}{2}*\sqrt{\frac{3a^2}{4}[/tex]
We can distribute the square root the division just like how we distributed the exponent 2, since the square root can be expressed as an exponent (1/2)
[tex]A = \frac{a}{2}*\frac{\sqrt{3a^2}}{\sqrt{4}}[/tex]
There's a radical identity that states: [tex]\sqrt[n]{a} * \sqrt[n]{b} = \sqrt[n]{a*b}[/tex]. We can use this to rewrite one radical as multiple radicals to simplify it:
[tex]A = \frac{a}{2}*\frac{\sqrt{a^2}*\sqrt{3}}{2}[/tex]
Simplify:
[tex]A = \frac{a}{2}*\frac{a*\sqrt{3}}{2}[/tex]
Now multiply the two fractions
[tex]A = \frac{a^2*\sqrt{3}}{4}[/tex]
This is the formula for the area of an equilateral triangle, but it is also often written as:
[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]
the cowboys football team started at the zero yard line.throughout the course of the first quarter the team gained 7 yards, lost 8 yards and then lost 7 more yard. what is the total of these three plays? how many yards did the team gain or lose?
The net number of yards lost by the cowboys football team is; 8 yards.
How many yards did the team gain or lose?It follows from the task content that the number of yards gained or lost by the football team during each of the three plays is; gained 7 yards, lost 8 yards and then lost 7 more yard.
Hence, by polarising yard gain as positive and yard loss as negative; we have;
Net total = 7 - 8 -7 = -8
Hence, the team lost 8 yards in the total game.
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Write the positive or negative number that best
represents the given information.
12 ft above sea level
Answer:
12
Step-by-step explanation:
Usually, above sea level is represented by a positive number and below sea level by a negative number.
12 ft above sea level = 12
Consider these two functions:
F(x) = 2cos(πx)
G(x) = 1/2cos(2x)
What are the amplitudes and periods of the two functions?
Answer:
[tex]\mathrm{pK}_{\mathrm{a}}+\mathrm{pK}_{\mathrm{b}}=\mathrm{pK}_{\mathrm{w}}[/tex] for [tex]\mathrm{HCl}[/tex] \& [tex]\mathrm{ClOH}[/tex]
Answer:
f(x): Amplitude = 2, Period = 2
g(x): Amplitude = ¹/₂, Period = π
Step-by-step explanation:
The cosine function is periodic, meaning it repeats forever.
Standard form of a cosine function:
f(x) = A cos(B(x + C)) + D
where:
A = amplitude (height from the mid-line to the peak)2π/B = period (horizontal distance between consecutive peaks)C = phase shift (horizontal shift - positive is to the left)D = vertical shiftFunction f(x)
[tex]f(x)=2 \cos (\pi x)[/tex]
Comparing this with the standard form of a cosine function:
Amplitude = 2[tex]\sf Period = \dfrac{2 \pi}{\pi}=2[/tex]Function g(x)
[tex]g(x)=\dfrac{1}{2} \cos (2x)[/tex]
Comparing this with the standard form of a cosine function:
[tex]\sf Amplitude=\dfrac{1}{2}[/tex][tex]\sf Period = \dfrac{2 \pi}{2}=\pi[/tex]Learn more about the cosine graph here:
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Find the circumference and the area of a circle with diameter 4 cm.
Use the value 3.14 for
, and do not round your answers. Be sure to include the correct units in your answers.
Answer:
Step-by-step explanation:
Since diameter x pi=3.14,this should be easy.
4x3=12
4x14/100=.56
12+.56=12.56
12.56 is the answer
To graph the inequality t>3, you would put an open circle on 3 and shade to the left
Answer:
It is an open circle and you would shade to the right on a number line.
Step-by-step explanation:
The circle is open because you are not including 3. Put numbers in for t and you will see that the numbers greater than 3 are all on the right side of the number line.
Integral of
11-3x/
x²+2x-3
dx
Answer:
2x+2
Step-by-step explanation:
x²+2x-3
=2(x^2-1)+2*1(2x^1-1)
What is the angle of rotation of the following figure?
This snow flake-like figure can be generated by rotating an end 60° five times around the center of the hexagon. There are two forms: (i) clockwise, (ii) counterclockwise.
What is the angle of rotation of a snow flake?
Geometrically speaking, snow flakes represent regular hexagons. Regular hexagons can divided into six concentric regular triangles, whose central angles have a measure of 60°. This fractal figure can be generated by rotating 60° five times around the center.
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The function h(x) is a transformation of the square root parent function,
f(t) = t. What function is H(x)?
Answer:
A. [tex]h(x)=\sqrt{x-3}[/tex]
Step-by-step explanation:
Step 1: DefinitionThe parent function of [tex]\sqrt{x}[/tex] is translated to the left when [tex]h[/tex] is positive in the transformation [tex]\sqrt{x+h}[/tex].
If [tex]h[/tex] is negative, the graph translates towards the left with the distance equal to the value of [tex]h[/tex].
Step 2: ImplementationHere the graph moved 3 units towards the right. This means that [tex]h[/tex] is negative and has the value of 3.
So, plugging that into the parent function for translation, the function becomes:
[tex]h(x)=\sqrt{x-3}[/tex]
Jakobe runs a coffee cart where he sells coffee for $1.50 tea for $2 , and donuts for $0.75 . On Monday, he sold 320 items and made $415 . He sold 3 times as much coffee as tea. How many donuts did he sell?
Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
Let x represent the amount of coffee, y represent the amount of tea and z represent the amount of donuts, hence:
x + y + z = 320 (1)
Also:
1.5x + 2y + 0.75z = 415 (2)
And:
x = 3y (3)
From the equations:
x = 150, y= 50 and z = 120
Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.
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Rationalize the denominator and simplify
√a+1-2/√a+1+2
5+a-4√(a+1)/(a-3) is the simplified form of the number. This can be obtained by multiplying numerator and denominator with conjugate of the denominator.
Simplify the number:Given the number,
√a+1-2/√a+1+2
Denominator = √a+1+2
Conjugate of the denominator = √a+1 - 2
Now, (√a+1-2)(√a+1 - 2 )/√(a+1+2)(√a+1 - 2 )
=(√a+1 - 2 )²/(√a+1)² - 4
=(a+1 -4√a+1+5)/(a+1-4)
=5+a-4√(a+1)/(a-3)
Hence 5+a-4√(a+1)/(a-3) is the simplified form of the number.
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When a constant force is applied to an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with mass 2kg, the acceleration of the object is 12 m/s^2. When the same force acts upon another object, its acceleration is 8m/s^2. What is the mass of this object?
The mass of the second object exists at 3 kg.
How to estimate the mass of the second object?Let, F be the Force
a be the acceleration
m be the mass
The formula of force exists F = m [tex]*[/tex] a
Since force exists constant, then:
[tex]$m_1 = 2 kg, a_1 = 12 m/s^2[/tex] subset 1 designates the first object
[tex]$m_2 = ? , a_2 = 8m/s^2[/tex] subset 2 designates the second object
To estimate the mass of the second object
A constant force indicates [tex]$ m_1 * a = m_2 * a[/tex]
The mass of second object [tex]m_2[/tex], we get
[tex]m_1 * a = m_2 * a[/tex]
[tex]$m_2 = \frac{m_1 * a}{a}[/tex]
[tex]$=\frac{2*12}{8}[/tex]
= 24/8 = 3 kg
Therefore, the mass of the second object exists at 3kg.
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During a flu epidemic a company with 200 employees had 1/3 on Monday and another 3/10 call in sick on Tuesday.
Part (a)
[tex]\frac{1}{10}+\frac{3}{10}=\frac{4}{10}=\boxed{\frac{2}{5}}[/tex]
Part (b)
[tex]200\left(\frac{2}{5} \right)=\boxed{80}[/tex]
The soil samples for the next field indicate that fertilizer
In order to have a greater fertilizer coverage, you need to increase flow rate and: A. increase speed to approximately 7.1 mph so that you cover the field more quickly.
What is a flow rate?A flow rate is an average rate and it can be defined as the number of flow units that are created per unit time such as fertilizer coverage on a farm.
In this context, we can infer and logically deduce that you have to increase flow rate and your increase speed to approximately 7.1 mph, so that you can cover the field more quickly and have a greater fertilizer coverage.
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Complete Question:
The soil samples for the next field indicate that fertilizer coverage needs to be greater. To achieve this, you need to increase flow rate. How would you achieve this?
A. Increase speed to approximately 7.1 mph so that you cover the field more quickly
B. Increase the engine speed to approximately 2,000 rpm
C. Decrease speed to approximately 6.0 mph so that you cover the field more slowly
D. Shift to second gear so that the engine speed slows
if Jill's age is 84 and johns age is 21 how many years ago was Jill 10 times older than john
1. In a class of 60 students, a survey was conducted, 30 students had applied for Addis Ababa University, 25 students applied for Bahir Dar University and 24 students applied for Wachemo University. 11 students applied for both Addis Ababa and Bahir Dar Universities, 6 applied for both Addis Ababa and Wachemo Universities, 9 applied for both Wachemo and Bahir Dar Universities while 4 applied neither of the aforementioned universities. Find 1.. number of students that applied for all the universities. 2 number of students that applied for at least two of the universities. 3 number of students that applied at most two universities. 4 number of students that applied for Addis Ababa but not Bahir Dar University? Please I need your help?
1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].
2. We have
[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]
Similarly we can find
[tex]n(A\cap B' \cap W) = 3[/tex]
[tex]n(A'\cap B\cap W) = 6[/tex]
[tex]n(A\cap B'\cap W') = 16[/tex]
[tex]n(A'\cap B\cap W') = 8[/tex]
[tex]n(A'\cap B'\cap W) = 12[/tex]
Then the total number of students that applied to at least two of the universities is
[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]
3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are
[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]
If we want students that applied to at least 1 school, we omit the first term and get a total of 36.
4. Split this set of students into those that also applied to Wachemo and those that did not.
[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]
I need help to this it would be awesome if you could help.
In the given diagram, the value of JI is 6
Solving equations
From the question, we are to determine JI
From the given diagram, we can write that
JI + IH = x + 22
Also,
JI = 2x + 28
and
IH = 5
Then, we can write that
2x + 28 + 5 = x + 22
Collect like terms
2x - x = 22 -28 -5
x = -11
Now, substitute the value of x into the equation
JI = 2x + 28
JI = 2(-11) + 28
JI = -22 +28
JI = 6
Hence, the value of JI is 6
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Question 1 of 10
If f(x)=x-1, which of the following is the inverse of f(x)?
O A. f¹(x) = x
OB. f¹(x)=x+1
O C. f¹(x)=x-1
O D. f¹(x) = 1 - x
SUBMIT
a two digit number has twice as many ones as tens twice the original number is 9 more than the reversed number find the original number
Answer: 36
Step-by-step explanation:
As this is a two-digit number, it will have a digit in its 10's place and a digit in its 1's place. We can represent the 10's digit using x and the 1's digit using y.
Since the ones is twice the number of tens, this means that y is equal to 2*x. Let's put this into an equation.
[tex]y=2x[/tex]
If the 10's digit is x and the 1's digit is y, the original number would be [tex]10x + y[/tex], as x must be multiplied by 10 to get it to the 10's place. Similarly, the reversed number would be [tex]10y + x[/tex], as y must be multiplied by 10 to get it to the 10's place.
Twice the original number (10x+y) is equal to 9 more than (i.e. 9 plus) the reversed number. We can put this into an equation to help us answer the question.
[tex]2(10x+y)=9+10y+x\\20x+2y=9+10y+x\\19x+2y=9+10y\\19x=9+8y[/tex]
Now we have a system of two equations.
[tex]y=2x\\19x=9+8y[/tex]
We can solve this system by substituting y for 2x in the second equation. Then, we can isolate and get the value of x.
[tex]19x=9+8(2x)\\19x=9+16x\\3x=9\\x=3[/tex]
Now that we have the value of x, let's put it back into the first equation and solve for y.
[tex]y=2(3)\\y=6[/tex]
Remember that x is the 10's digit and y is the one's digit of our answer. Since x is 3 and y is 6, our answer is 36.
CheckingWe can quickly check if our answer is right by making sure both conditions in the question are met.
6 (the one's digit) is twice the value of 3 (the ten's digit), making the first condition true. Twice of 36 is 72, which is 9 more than the reversed number, which is 63.
A cube with an edge of 10cm is filled with water.How much can it contain?
The cube can contain 1000 cm^3 of water
How to determine the amount of water?The side length is given as:
L = 10 cm
The amount of water it can contain is calculated as;
V =L^3
So, we have:
V = 10^3
Evaluate
V = 1000
Hence, the cube can contain 1000 cm^3 of water
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-4
-2
4
-2-
0
3
On which interval are both functions increasing?
2
-8
p
0
-X
-2
g(x) = -x - 3| +4
(,0)
Answer: [tex](3, \infty)[/tex]
Step-by-step explanation:
q(x) is increasing for [tex]x > 3[/tex].
p(x) is increasing for [tex]x \geq -2[/tex].
So, both functions are increasing for [tex](3, \infty)[/tex].
A trapezoid has an area given by A =1-2(b1-b2)h,where b1,is the length of one base, b2 is the length of the other
base, and h is the height. If the trapezoid has an area of 14 square units, solve for b in terms of the other variables.
The value of [tex]b_{1}[/tex] is 69/4h and value of [tex]b_{2}[/tex] is 43/4h if the area of trapezoid is 14 square units.
Given that area is equal to 1-2([tex]b_{1} -b_{2}[/tex])h and area is 14 square units.
We know that area of trapezoid is equal to (a+b)h/2.
We have to put the given equation equal to 14 to get our first equation.
14=1-2([tex]b_{1} -b_{2}[/tex])h
14=1-2[tex]b_{1}[/tex]h+2[tex]b_{2}[/tex]h
2[tex]b_{1}[/tex]h-2[tex]b_{2}[/tex]h=-13------------1
Now put the value of area in the formula of area of trapezoid.
14=[tex](b_{1} +b_{2})h/2[/tex]
[tex]b_{1}[/tex]h+[tex]b_{2}h[/tex]=28-----------------2
Multiply equation 2 by 2 and then subtract 2 from 1
[tex]2b_{1}h -2b_{2}h -2b_{1}h-2b_{2}h[/tex]=-13-56
-4[tex]b_{2}h[/tex]=-43
[tex]b_{2}[/tex]=43/4h
Put the value of [tex]b_{2}[/tex] in 2 to get the value of [tex]b_{1}[/tex].
[tex]b_{1} h+43/4h*h=28[/tex]
[tex]b_{1}[/tex]=69/4h
Hence The value of [tex]b_{1}[/tex] is 69/4h and value of [tex]b_{2}[/tex] is 51/4h if the area of trapezoid is 14 square units.
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A washer and a dryer cost $737 combined. The washer costs $37 more than the dryer. What is the cost of the dryer
Answer: $350
Step-by-step explanation:
$700 divided by 2 is 350
You have customers of the following ages, what is the median age of your customer base? 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70
The median age of your customer base = 45
What is median?The median is the value that divides a data sample, a population, or even a probability distribution's upper and lower halves in statistics and probability theory. It could be referred to as "the middle" value for a data set. The fundamental difference between the mean and the median when describing data is that the median is more representative of the "typical" value because it is not skewed by the a small proportion of extremely large or small values. Due to the fact that income distribution can be highly skewed, the median income, for instance, might be a better indicator of what is considered a "typical" income.
Mean = middle term
= 6th term
= 45
Hence, Median of age = 45.
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help me pleaseeeee (20 points)
what is the volume of this solid?
Answer:
A) 39
Step-by-step explanation:
If you calculate the volume of the cube which is the V=a^3 you get 27, then calculating the pyramid, the Length is 3, and the Width is also 3. Then the height is 4. so you put that into the formula to calculate the volume which is V=L*W*H
--------
3
you get 12. Now its simple, you just add the 2 volumes, so you do 27+12 and its 39.
39 is the answer by calculating the volume using formula!
I’m confused in the exponential
Answer:
this mean in case of multiplying exponents with same base you should add the exponent and keep base same.
example:
[tex] {6}^{4} \times {6}^{5 } = {6}^{4 + 5} \\ = {6}^{9} [/tex]