The set-builder form is { x | x is a calculus book}.
A set is a collection of quantities based on a specific criteria. A set can be expressed in two forms.
We usually use the form {a, b, c,....} in which we include each element of the set separated with a comma in between. This form of set notation is called the roster form.
Another set notation is the set-builder form. This form is more concise and precise and easily understandable also. In this form, all the elements are included in the set which has a same property. That is, here we don't write each and every element separately in the set but just emphasize the property of the elements which are included in the set.
We are asked to write the set of all calculus books in set-builder form.
The set-builder form for 'The set of all calculus books' is { x | x is a calculus book}.
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Astronomers oftenmeasure largedistances usingastronomical units(AU) where 1 AU isthe average distancefrom Earth to theSun. In the image, drepresents thedistance from a starto the Sun. Using atechnique called"stellar parallax,astronomers determinedO is 0.00001389degrees.1. How faraway isthe starfrom theSun inastronomical units?Showyourreasoning.2. Write anexpression tocalculated for anystar.
In the given right triangle of the figure
we have that
sin(O)=1/d ----> by the opposite side divided by the hypotenuse in a right triangle
solve for d
d=1/sin(O)
[tex]d=\frac{1}{sin\left(0.00001389^o\right?}[/tex]d=4,124,966.13 AU -----> two decimal places
the expression to calculate d for any star is equal to
[tex]d=\frac{1}{sin\left(O\right)}[/tex]Select True or False for each statement.
A right triangle always has obtuse exterior angles at two vertices.
Answer:
true
Step-by-step explanation:
it is true ........
Explanation:
A right triangle has exactly one angle that is 90 degrees. This is called a right angle.
The other two angles are acute, which means they are less than 90 degrees. An example would be 30 degrees and 60 degrees.
If 30 degrees is an interior acute angle, then 180-30 = 150 degrees is the exterior obtuse angle. Similarly, the adjacent angle to the 60 is 180-60 = 120 degrees.
This example shows we have two obtuse exterior angles. This applies to any right triangle, and not just this particular one.
Solve the problems.
Prove: BD = CD
The Angle-Side-Angle (ASA) criterion states that any two angles and the side included between them of one triangle are identical to the corresponding angles and the included side of the other triangle if two triangles are congruent. One of the requirements for two triangles to be congruent is angle side angle.
When two parallel lines are intersected by another line, comparable angles are the angles that are created in matching corners or corresponding corners with the transversal.
If the three sides and the three angles of both angles are equal in any orientation, two triangles are said to be congruent.
Given,
M∠1 = M∠2
(On joining BD and CD)
M∠ADB = M∠ADC
In ΔABD and ΔADC
M∠ADB = M∠ADC (Given)
M∠BAD = M∠DAC (Given)
AD = AD (Common Sides)
⇒ ΔABD ≅ ΔADC (Angle Side Angle Property)
So, BD = CD (Corresponding sides are equal to a Congruent Triangle)
Hence, proved that BD = CD.
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The height of a tree increases as time passes. Your friend says that time is the dependent variable because size depends on time. Is your friend correct?
Given that the height of a tree increases as time passes.
Your friend says that is dependent because size deemds on time.
Leet's determine if your friend is correct.
Here, since the height of a tree increases as time passes, the height of the tree depends on the time.
Hence, size is the dependent variable while time is the independent variable.
Therefore, your friend is NOT correct.
ANSWER:
No, your friend is NOT correct.
If XD = 2X – 6 and XV = 3x – 6 and WY and XV bisect at D, what is XV?
The diagonal WY bisects the diagonal XV at point D, which means that XV is divided into two equal line segments XD and XV.
[tex]XD=XV[/tex]Replace the equation above with the given expressions for both line segments:
XD= 2x-6
XV= 3x-6
[tex]2x-6=3x-6[/tex][tex]undefined[/tex]helppppppppppppp meeeeeeeeeeeeeeeeeeeeeeeeeee
Answer:
55.17
Step-by-step explanation:
[tex]P(0)=0.023(0)^3-0.289(0)^2+3.068(0)+55.170=55.17[/tex]
Out of 450 applicants for a job, 249 are female and 59 are female and have a graduate degree. Step 1 of 2 : What is the probability that a randomly chosen applicant has a graduate degree, given that they are female? Express your answer as a fraction or a decimal rounded to four decimal places.
The probability that a randomly chosen applicant has a graduate degree, given that they are female is 0.236.
What is the probability?Probability refers to a possibility that deals with the occurrence of random events. The probability of all the events occurring need to be 1.
P(E) = Number of favorable outcomes / total number of outcomes
Given that 249 are female and 59 are female and have a graduate degree out of 450 applicants.
We are given following in the question;
M: Applicant is male.
G: Applicant have a graduate degree
F : Applicant is female.
The Total number of applicants = 450
Number of female applicants = 249
Number of female applicants have a graduate degree = 59
Therefore,
P(G/F) = P([tex]G^F[/tex])/P(F)
= 59/207 or 0.236
Hence, the probability that a randomly chosen applicant has a graduate degree, given that they are female is 0.236.
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please di it quickly I just need to confirm answer
The Solution:
Given:
[tex]\begin{gathered} V=(-5,3) \\ \\ W=(\frac{3}{2},-\frac{1}{2}) \end{gathered}[/tex]Required:
To find the value of V - W
[tex]V-W=\lbrace(-5-\frac{3}{2}),(3--\frac{1}{2})\rbrace=(-6\frac{1}{2},3\frac{1}{2})=(-\frac{13}{2},\frac{7}{2})[/tex]
Therefore, the correct answer is [option 3]
the bearing of L from Q is 90° what is the bearing of Q for L
Given:
the bearing of L from Q is 90°
Required:
what is the bearing of Q for L
Explanation:
There is a 180 degree difference in bearing between two location(L from Q, Q from L)
If L from Q is 90 degree then
Q from L is
180-90=90degree
Required answer:
90 degree
I'm behind in geometry if you could teach me this I would highly appreciate it
We know that AC and BD bisect each other, but AC is not equal to BD.
Basically, the problem is saying that these segments are bisectors of each other, that can be represented as the image below shows
As you can see in the image, AC and BD bisect each other, but their lengths are not equal.
Find the midpoint of AB given A (-3,-5) and B (-1,8)
A: (-3,-5)
B: (-1,8)
To find the midpoint we have to add each coordinate and divide by 2:
Midpoint: (-3-1)/2 , (-5+8)/2 = -4/2, 3/2 = -2,1.5
Midpoint: (-2, 1.5)
Luther made $9,000 in interest by placing $60,000 in a savings account with simple interest for 3 years. What was the interest rate?
Answer:
[tex]5\%\text{ or }0.05[/tex]
Step-by-step explanation:
Simple interest rate means if I have "P" amount that I initially deposited, then every year this amount increases by "x% of P" or the original amount I deposited. So it's increasing by the same amount each year, unlike compound interest.
The formula for calculating the amount of interest is: [tex]I=P*r*t[/tex]
Where, r is the interest rate, t is the time unit, and P is the initial amount.
In most cases the t will be expressed in years, and one thing to note is the r is the interest rate in decimal form, so [tex]30\%=0.30[/tex], we want to convert it to decimal form by dividing by 100
We know the amount of interest, as it's given to us as 9,000, and the principle amount or initial amount, given to us as 60,000, and also the time which is given to us as 3 years.
So we know that:
[tex]I=9,000\\P=60,000\\t=3\\[/tex]
Plugging all these values into the equation we get:
[tex]9,000=60,000*3*r\\\\9,000=180,000*r\\\\\frac{9,000}{180,000}=r\\\\0.05=r[/tex]
as noted above, this interest rate, "r" is expressed in decimal form. Since we have to divide by 100 to convert from percentage to decimal, we have to multiply by 100 to convert from decimal to percentage.
this gives us: [tex]r=5\%[/tex]
0.04 divided by 0.628
Answer:
0.1 also 0.0636942675
A ship leaves port on a bearing of 34.0 degrees and travels 10.4 mi. The ship then turns due east and travels 3.9 mi. How far is the ship from port, and what is its bearing from port?
→
r
=
13.5
mi
, at a bearing angle of
50.4
o
Explanation:
We're asked to find the total displacement, both the magnitude and direction, of the ship after it leaves the port with the given conditions.
First, I'll explain what a bearing is.
A bearing is NOT a regular angle measure; normally, angles are measured anticlockwise from the positive
x
-axis, but bearing angles are measured clockwise from the positive
y
-axis.
Therefore, a bearing of
34.0
o
indicates that this is an angle
90.0
o
−
34.0
o
=
56.0
o
measured normally. We'll use this angle in our calculations.
We're given that the first displacement is
10.4
mi
at an angle of
56.0
o
(as calculated earlier). Let's split this up into its components:
x
1
=
10.4
cos
56.0
o
=
5.82
m
y
1
=
10.4
sin
56.0
o
=
8.62
m
Our second displacement is a simple
4.6
mi
due east, that is, the positive
x
-direction. The components are thus
x
2
=
4.6
mi
y
2
=
0
mi
To find the total displacement from the port, we'll add these two vectors' components and use the distance formula:
Δ
x
=
x
1
+
x
2
=
5.82
mi
+
4.6
mi
=
10.42
mi
Δ
y
=
y
1
+
y
2
=
8.62
mi
+
0
mi
=
8.62
mi
r
=
√
(
x
total
)
2
+
(
y
total
)
2
=
√
(
10.42
mi
)
2
+
(
8.62
mi
)
2
=
13.5
mi
The direction of the displacement vector is given by
tan
θ
=
Δ
y
Δ
x
so the angle is then
θ
=
arctan
(
Δ
y
Δ
x
)
=
arctan
(
8.62
mi
10.42
mi
)
=
39.6
o
The question asked for the bearing angle, which is just this angle subtracted from
90
o
:
Bearing angle
=
90
o
−
39.6
o
=
50.4
o
Which equation is true when A.n = 1.2 B.
1.2n=10 n+1=1.2 C. 5+n=6.2 D. 10n=1.2
The true equation is 5 + n = 6.2.
Here we have to find the equation for which n = 1.2.
So the first equation is
10n = 1.2
So for this, we get the value of n as:
n = 1.2/10
= 0.12
which is not equal to 1.2.
So it is not correct
The second equation is:
n + 1 = 1.2
n = 0.2
which is not equal to 1.2
So it is also not correct.
The third equation is:
5 + n = 6.2
n = 6.2 - 5
= 1.2
So it is correct.
The fourth equation is:
1.2n = 10
n = 10/1.2
= 8.33
Here n is not equal to 1.2
So it is also correct.
Therefore the correct equation is 5 + n = 6.2.
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7 Write the numbers in order from
GREATEST to LEAST.
-71.1 -71 1/2
How long is the control line? I couldn’t figure this out
Solution:
Given the circle with center A as shown below:
The plane travels 120 feet counterclockwise from B to C, thus forming an arc AC.
The length of the arc AC is expressed as
[tex]\begin{gathered} L=\frac{\theta}{360}\times2\pi r \\ \text{where} \\ \theta\Rightarrow angle\text{ (in degre}e)\text{subtended at the center of the circle} \\ r\Rightarrow radius\text{ of the circle, which is the }length\text{ of the control line} \\ L\Rightarrow length\text{ of the arc AC} \end{gathered}[/tex]Given that
[tex]\begin{gathered} L=120\text{ f}eet \\ \theta=80\degree \\ \end{gathered}[/tex]we have
[tex]\begin{gathered} L=\frac{\theta}{360}\times2\pi r \\ 120=\frac{80}{360}\times2\times\pi\times r \\ cross\text{ multiply} \\ 120\times360=80\times2\times\pi\times r \\ \text{make r the subject of the equation} \\ \Rightarrow r=\frac{120\times360}{2\times\pi\times80} \\ r=85.94366927\text{ fe}et \end{gathered}[/tex]Hence, the length of the control line is 85.94366927 feet.
Hello me with part B pleaseeee
Answer:
0, 0, 0
Step-by-step explanation:
You want the sum of a number and its opposite for the numbers ...
3, 7.5, and -2 2/3
Additive inverseThe definition of the additive inverse (opposite) of a number is that it is the number that produces 0 when summed with the original number.
Any number summed with its opposite will give zero.
The sums are ...
3 + (-3) = 07.5 + (-7.5) = 0-2 2/3 + (2 2/3) = 0What is x2 + 6x complete the square
You have the following expression:
x² + 6x
In order to complete the square, take into account that 6 is two times the product of the first coeffcient by the second one in the binomial (a+ b)², then, you have:
6 = 2ab
a=1 because is the coeffcient of the term with x², then for b you obtain:
b = 6/2(1) = 3
the third term of the trynomial is the squared of b.
Evaluate. 12⋅(1/4+1/3)to the power of2+2/3 Enter your answer as a mixed number in simplest form by filling in the boxes.
12×(1/4+1/3)to the power of2+2/3 using PEDMASand INDICIES rule gives 7^8/3
What is Indices? lndicies is expressed as Ax^n. Where A is the coefficient, x is the base and n is the power or index.
12×(1/4+1/3)to the power of2+2/3
Evaluating the expression
= (12×( 1/4+1/3))^2+2/3
using PEDMAS
= (12× 7/12)^8/3
opening the bracket, we therefore have
= 7^8/3
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Simplify the trigonometric expression. cos(theta+pi/2)
We have to simplify the expression:
[tex]\cos (\theta+\frac{\pi}{2})[/tex]We could see it graphically:
We see that for any angle theta, the cosine of theta + pi/2 is equal to negative sin of theta.
Then we can write:
[tex]\cos (\theta+\frac{\pi}{2})=-\sin (\theta)[/tex]The answer is -sin(theta).
Function A gives the audience in millions
Using function concepts, it is found that:
a) The meaning of each expression is given as follows:
A(4) = audience after four hours.A(0.5) = 1.5 = the audience after 0.5 hours is of 1.5 million peopleb) The expression is: A(4) = 1.3.
c) The expression is: A(2) = A(2.5).
FunctionIn the context of this problem, the format of the function is:
A(t).
In which the meaning of each variable is given as follows:
t is the time in hours after the beginning of the show.A(t) is the audience, in millions of hours.Which gives the meaning of each expression in item a.
For item b, the expression is given as follows:
A(4) = 1.3.
As 4 hours after the episode premiered, the audience was of 1.3 million people.
For item c, the expression is given as follows:
A(2) = A(2.5).
As the audience after 2 hours = 120 minutes is the same as the audience half an hour = 30 minutes later.
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The side of a square is represented by (s - 4);1. Write an expression that could represent the perimeter of the square.2. If the perimeter of the square is no more than 80 ft, what is the maximum value of s?
We have a square with sides (s-4).
The perimeter of a square is 4 times the length of the side, so we can write:
[tex]P=4\cdot(s-4)=4s-16[/tex]If P is no more than 80 ft, we can find the maximum value for s as:
[tex]\begin{gathered} P<80 \\ 4s-16<80 \\ 4s<80+16 \\ 4s<96 \\ s<\frac{96}{4} \\ s<24 \end{gathered}[/tex]Answer:
1) The perimeter is P = 4s-16
2) The maximum value for s, if the perimeter is no more than 80 ft, is 24 ft.
16. The height, h, in feet of an object above the ground is given by h = -16t² +64t+190, t≥0, where t is the time in seconds. a) b) c) d) When will the object be 218 feet above the ground? When will it strike the ground? Will the object reach a height of 300 feet above the ground? Find the maximum height of the object and the time it will take.
the maximum height is 254 feet.
Answer:
a) 0.5 seconds and 3.5 seconds.
b) 5.98 seconds (2 d.p.)
c) No.
d) 254 feet at 2 seconds.
Step-by-step explanation:
Given equation:
[tex]h=-16t^2+64t+190, \quad t \geq 0[/tex]
where:
h is the height (in feet).t is the time (in seconds).Part aTo calculate when the object will be 218 feet above the ground, substitute h = 218 into the equation and solve for t:
[tex]\begin{aligned}\implies -16t^2+64t+190 & = 218\\-16t^2+64t+190-218& = 0\\-16t^2+64t-28 & = 0\\-4(4t^2-16t+7) & = 0\\4t^2-16t+7 & = 0\\4t^2-14t-2t+7 &=0\\2t(2t-7)-1(2t-7)&=0\\(2t-1)(2t-7)&=0\\\implies 2t-1&=0\implies t=\dfrac{1}{2}\\\implies 2t-7&=0 \implies t=\dfrac{7}{2}\end{aligned}[/tex]
Therefore, the object will be 218 feet about the ground at 0.5 seconds and 3.5 seconds.
Part bThe object strikes the ground when h is zero. Therefore, substitute h = 0 into the equation and solve for t:
[tex]\begin{aligned}\implies -16t^2+64t+190 & = 0\\-2(8t^2-32t-95) & = 0\\8t^2-32t-95 & = 0\end{aligned}[/tex]
Use the quadratic formula to solve for t:
[tex]\implies t=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]
[tex]\implies t=\dfrac{-(-32) \pm \sqrt{(-32)^2-4(8)(-95)} }{2(8)}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{1024+3040} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{4064} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{16 \cdot 254} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{16} \sqrt{254} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm 4 \sqrt{254} }{16}[/tex]
[tex]\implies t=\dfrac{8\pm \sqrt{254} }{4}[/tex]
As t ≥ 0,
[tex]\implies t=\dfrac{8+ \sqrt{254} }{4}\quad \sf only.[/tex]
[tex]\implies t=5.98 \sf \; s \; (2 d.p.)[/tex]
Therefore, the object strikes the ground at 5.98 seconds (2 d.p.).
Part c
To find if the object will reach a height of 300 feet above the ground, substitute h = 300 into the equation and solve for t:
[tex]\begin{aligned}\implies -16t^2+64t+190 & = 300\\-16t^2+64t+190-300 & =0\\-16t^2+64t-110 & =0\\-2(8t^2-32t+55) & =0\\8t^2-32t+55& =0\end{aligned}[/tex]
Use the quadratic formula to solve for t:
[tex]\implies t=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]
[tex]\implies t=\dfrac{-(-32) \pm \sqrt{(-32)^2-4(8)(55)} }{2(8)}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{1024-1760} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{-736} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{16 \cdot -1 \cdot 46} }{16}[/tex]
[tex]\implies t=\dfrac{32 \pm \sqrt{16} \sqrt{-1} \sqrt{ 46}}{16}[/tex]
[tex]\implies t=\dfrac{32 \pm 4i\sqrt{ 46} }{16}[/tex]
[tex]\implies t=\dfrac{8\pm \sqrt{ 46} \;i}{4}[/tex]
Therefore, as t is a complex number, the object will not reach a height of 300 feet.
Part dThe maximum height the object can reach is the y-coordinate of the vertex.
Find the x-coordinate of the vertex and substitute this into the equation to find the maximum height.
[tex]\textsf{$x$-coordinate of the vertex}: \quad x=-\dfrac{b}{2a}[/tex]
[tex]\implies \textsf{$x$-coordinate of the vertex}=-\dfrac{64}{2(-16)}=-\dfrac{64}{-32}=2[/tex]
Substitute t = 2 into the equation:
[tex]\begin{aligned}t=2 \implies h(2)&=-16(2)^2+64(2)+190\\&=-16(4)+128+190\\&=-64+128+190\\&=64+190\\&=254\end{aligned}[/tex]
Therefore, the maximum height of the object is 254 feet.
It takes 2 seconds for the object to reach its maximum height.
Consider the following equation of the circle. Graph the circle
Explanation
Given the equation;
[tex](x+6)^2+(y+7)^2=4[/tex]Using a graphing calculator, the graph of the circle becomes;
Answer:
someone please help…
Q.12 The polynomials are 2 and 3
What is polynomial ?
A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and positive-integer powers of variables. An example of a polynomial of a single indeterminate x is x² − 4x + 7.
Given, p(x) = -x² + x + 2
p has a degree 2
Let, a = 1 + i√2
b = 1 - i√2
a + b = 1 + i√2 + 1 - i√2
= 2
a * b = (1 + i√2) * (1 - i√2 )
= 1 - (i√2)²
= 1 - i²*2 (where, i² = -1)
= 1 + 2
= 3
Therefore, the polynomial are 2 and 3
Q.13 The polynomial is x³ + x
Given, Q(x) = x³ - 2x² - 1
Q has a degree of 3
=>(x - 0) (x - i) (x + i)
=>(x² - ix) (x + i)
=>x³ + x²i - ix² - i²x (where i² = -1)
=>x³ + x
Therefore, the polynomial is x³ + x
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What is the equation of a circle with radius 2 and center (3, 1)? (x - 3)2 – (y - 1)2 = 4 O (c - 3)2 + (y - 1)2 = 4 (x ) O (x + 3)2 + (y + 1)2 = 4 (x - 3)² + (y - 1)2 = 2
Given the radius of a circle as r = 2 and centre = (3,1)
We want to find the equation
Solution
First, the equation of a circle centre (a,b) and radius r is given by
[tex](x-a)^2+(y-b)^2=r^2[/tex]From the question
[tex]\begin{gathered} a=3 \\ b=1 \\ r=2 \end{gathered}[/tex]We put the parameters into the equation ans simplify
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-3)^2+(y-1)^2=2^2 \\ (x-3)^2+(y-1)^2=4 \end{gathered}[/tex]Therefore the correct option is Option B
I'll give brainliest!
Answer:
24
Step-by-step explanation:
8y^0 + 2y^2 * x^-1
8(4)^0 + 2(4)^2 * 2^-1
8 + 2 * 16 * 2^-1
8 + 2 * 2^-1 * 16
8 + 2^1 - 1 * 16
8 + 16
24
Hope this helps! :)
How do you find the square root of 18? Needs to be in decimal form and we cannot use calculators
Given:
[tex]\sqrt{18}[/tex]To find:
The root of 18 without using a calculator
There is a formula that gives an approximation of a square root without a calculator. This is given as:
[tex]\begin{gathered} \sqrt{N}\text{ = }\frac{N\text{ + M}}{2\sqrt{M}} \\ where\text{ N = is the number we want to find its root} \\ M\text{ = is a perfect square close the number we are to find} \\ That\text{ is a number we can find its root } \end{gathered}[/tex][tex]\begin{gathered} \text{N = 18} \\ M\text{ = 16 is the closest number to 18 we can find its root} \\ \\ substitute\text{ the values into the formula:} \\ \sqrt{18}\text{ = }\frac{18\text{ + 16}}{2\times\sqrt{16}} \\ \\ \sqrt{18}\text{ = }\frac{34}{2(4)} \end{gathered}[/tex][tex]\begin{gathered} \sqrt{18}\text{ = }\frac{34}{8} \\ \\ \sqrt{18}\text{ = 4.25} \end{gathered}[/tex]We are working on integers, and I just don’t get the concept, if you could do an EXAMPLE problem for me that would be great I understandan this I just need further instruction
The history of integers begins with the natural numbers (1,2,3,...), which are also called the counting numbers. They arise naturally by the need of counting things. But there is a problem. "There is a missing number". If you have a number n of things and you subtract all of them, what remains? Intuitively, it must be a number, a number representing nothing (to have nothing). This is the origin of zero.
Now, as time passes, the man began to do trades and arose the money. This leads to a problem. Other numbers are missing now. How do you express how much money you owe someone? counting numbers are not useful here, for they mean "the money you have". Th