Write the equation of the sinusoidal function shown?
A) y = cos x + 2
B) y = cos(3x) + 2
C) y = sin x + 2
D) y = sin(3x) + 2
Answer:
günah(3x) + 2
Step-by-step explanation:
Gösterilen sinüzoidal fonksiyonun denklemini yazınız? A) y = cos x + 2 B) y = cos(3x) + 2 C) y = günah x + 2 D) y =
Answer:
y = sin(3x) + 2
what does this equal 2^3 + 6^5=
[tex]\\ \sf\longmapsto 2^3+6^5[/tex]
[tex]\\ \sf\longmapsto 2^3+(2\times 3)^5[/tex]
[tex]\\ \sf\longmapsto 8+2^5\times 3^5[/tex]
[tex]\\ \sf\longmapsto 8+32\times 243[/tex]
[tex]\\ \sf\longmapsto 40+7776[/tex]
[tex]\\ \sf\longmapsto 7784[/tex]
Answer:
2*2*2= 8
6*6*6*6*6= 7,776
7,776+8=
7,784
Help me plezzzzzzzzzzzzzzzzz
Answer:
Step-by-step explanation:
I am assuming 1 and 2 are asking for factors,
I am only gonna solve 4 for the sake of time,
9x^5-x^3+2x^2-x
x(9x^4-x^2+2x-1)
x(9x^4-(x^2-2x+1)
x(9x^4-(x-1)^2)
x(3x^2+x-1)(3x^2-x+1)
x^4+2x^2-24
x^4+6x^2-4x^2-24
x^2(x^2+6)-4(x^2+6)
(x^2-4)(x^2+6)
(x+2)(x-2)(x^2+6)
a^4+a^2+1
a^4+1+a^2
(a^2+1)^2-2a^2+a^2
(a^2+1)^2-a^2
(a^2+1-a)(a^2+1+a)
(a^2+a+1)(a^2-a+1)
a^3+b^3+c^3-3abc
=(a+b)^3+c^3−3ab(a+b)−3abc
=(a+b+c)^3−(3c(a+b)^2+3(a+b)c^2)−3ab(a+b+c)
=(a+b+c)^3−3c(a+b)(a+b+c)−3ab(a+b+c)
=(a+b+c)^3−(a+b+c)(3ab+3bc+3ac)
=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac)−(a+b+c)(3ab+3bc+3ac)
=(a+b+c)(a^2+b^2+c^2−ab−bc−ac)
Determine the sum of the first 33 terms of the following series:
−52+(−46)+(−40)+...
Answer:
1320
Step-by-step explanation:
Use the formula for sum of series, s(a) = n/2(2a + (n-1)d)
The terms increase by 6, so d is 6
a is the first term, -56
n is the terms you want to find, 33
Plug in the numbers, 33/2 (2(-56)+(32)6)
Simplify into 33(80)/2 and you get 1320
Rationalise the denominator
Answer:
sqrt(3) /3
Step-by-step explanation:
1 / sqrt(3)
Multiply the top and bottom by sqrt(3)
1/ sqrt(3) * sqrt(3)/ sqrt(3)
sqrt(3) / sqrt(3)*sqrt(3)
sqrt(3) /3
Answer:
[tex] = { \sf{ \frac{1}{ \sqrt{3} } }} \\ \\ { \sf{ = \frac{1}{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} } }} \\ \\ = { \sf{ \frac{ \sqrt{3} }{ {( \sqrt{3}) }^{2} } = \frac{ \sqrt{3} }{3} }} [/tex]
What is the range of the absolute value function shown in the graph?
A. 3 ≤ y < ∞
B. -∞ < y ≤ 3
C. -6 ≤ y < ∞
D. -∞ < y < ∞
Answer:
C
Step-by-step explanation:
as we can see on the graph, the lowest y value is the vertex/corner at x=3, y=-6.
all other y values are above (=are larger) that level.
and it goes up without appearant limit, so up to infinity.
help me pls??????? :)
Answer:4 in each bad 2 left over
Step-by-step explanation:
Answer:
4 in each bag and 2 left over
Step-by-step explanation:
divide 14 by 3
3 goes into 14, 4 times
14 - 12 = 2
4 in each bag and then 2 left over
What is the volume of a sphere with a diameter of 7.7 ft, rounded to the nearest tenth
of a cubic foot?
Step-by-step explanation:
V=4/3πr^3
V=4/3π(3.85)^3
V=4/3π(57.066625)
V=4/3 (179.280089865)
V=239.04011982
V=239 ft^3
Solve for a. HELPPppppppppppppppp
Answer:
a = -24
Step-by-step explanation:
[tex]30=26-\frac{a}{6}\\(6)4=-\frac{a}{6}(6)\\24=-a\\-24=a[/tex]
Answer:
-24 = a
Step-by-step explanation:
30 - 26 - a/6
Subtract 26 from each side
30 -26 = 26 - a/6 - 26
4 = -a/6
Multiply by -6 on each side
-6 *4 = -a/6 *-6
-24 = a
Two observers are 300 ft apart on opposite sides of a flagpole. The angles of
elevation from the observers to the top of the pole are 20°
and 15°. Find the
height of the flagpole.
ax^2-y^2-x-y factorize
Answer:
x(ax-1)-y(y+1)
Step-by-step explanation:
you have to group the like terms
ax^2-x-y^2-y
x(ax-1)-y(y+1)
I hope this helps
100 POINTS AND BRAINLIEST FOR THIS WHOLE SEGMENT
a) Find zw, Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
b) Find z^10. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
c) Find z/w. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
d) Find the three cube roots of z in complex form. Give answers correct to 4 decimal
places.
Answer:
See Below (Boxed Solutions).
Step-by-step explanation:
We are given the two complex numbers:
[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]
First, convert z to polar form. Recall that polar form of a complex number is:
[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]
We will first find its modulus r, which is given by:
[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]
In this case, a = √3 and b = -1. Thus, the modulus is:
[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]
Next, find the argument θ in [0, 2π). Recall that:
[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]
Therefore:
[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]
Evaluate:
[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]
Since z must be in QIV, using reference angles, the argument will be:
[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]
Therefore, z in polar form is:
[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]
Part A)
Recall that when multiplying two complex numbers z and w:
[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]
Therefore:
[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]
Simplify. Hence, our polar form is:
[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]
To find the complex form, evaluate:
[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]
Part B)
Recall that when raising a complex number to an exponent n:
[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]
Therefore:
[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]
Substitute:
[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]
Simplify:
[tex]\displaystyle z^{10} = 1024\left(\cos\frac{55\pi}{3}+i\sin \frac{55\pi}{3}\right)[/tex]Simplify using coterminal angles. Thus, the polar form is:
[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]
And the complex form is:
[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]
Part C)
Recall that:
[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]
Therefore:
[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]
Simplify. Hence, our polar form is:
[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]
And the complex form is:
[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]
Part D)
Let a be a cube root of z. Then by definition:
[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]
From the property in Part B, we know that:
[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]
Therefore:
[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]
If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:
[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]
The first equation can be easily solved:
[tex]r=\sqrt[3]{2}[/tex]
For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:
[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]
Solve for the argument:
[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]
There are three distinct solutions within [0, 2π):
[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]
Hence, the three roots are:
[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]
Or, approximately:
[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]
−30=5(x+1)
what is x?
[tex]\\ \rm\Rrightarrow -30=5(x+1)[/tex]
[tex]\\ \rm\Rrightarrow -30=5x+5[/tex]
[tex]\\ \rm\Rrightarrow 5x=-30-5[/tex]
[tex]\\ \rm\Rrightarrow 5x=-35[/tex]
[tex]\\ \rm\Rrightarrow x=\dfrac{-35}{-5}[/tex]
[tex]\\ \rm\Rrightarrow x=7[/tex]
Answer:
x = -7
Step-by-step explanation:
-30 = 5 (x -1 )
5 ( x + 1 ) =-30
5 (x + 1 ) = - 30
5 5
x + 1 = -6
x + 1 -1 = -6 -1
x = - 7
Find the value of the sum 219+226+233+⋯+2018.
Assume that the terms of the sum form an arithmetic series.
Give the exact value as your answer, do not round.
Answer:
228573
Step-by-step explanation:
a = 219 (first term)
an = 2018 (last term)
Sn->Sum of n terms
Sn=n/2(a + an) [Where n is no. of terms] -> eq 1
To find number of terms,
an = a + (n-1)d [d->Common Difference] -> eq 2
d= 226-219 = 7
=> d=7
Substituting in eq 2,
2018 = 219 + (n-1)(7)
1799 = (n-1)(7)
1799 = 7n-7
1799 = 7(n-1)
1799/7 = n-1
257 = n-1
n=258
Substituting values in eq 1,
Sn = 258/2(219+2018)
= 129(2237)
= 228573
convert 10.09% to a decimal
Answer:
0.1009
Step-by-step explanation:
To convert percentage into decimal, you need to divide the percentage by 100
10.09/100
= 0.1009
Determine three consecutive odd integers whose sum is 2097.
Answer:
first odd integer=x
second odd integer=x+2
third odd integer=x+4
x+x+2+x+4=2097
x+x+x+2+4=2097
3x+6=2097
3x=2097-6
3x=2091
3x/3=2091/3
x=697
therefore, x=697
x+2=697+2=699
x+4=697+4=701
Which value of x makes this equation true?-9x+15=3(2-x)
Step-by-step explanation:
-9x+15=3(2-x)
expand the bracket by the right hand side6-6x
2. collect like terms
-9x+15= 6-6x
15-6 = 6x+9x
11= 15x
3. divide both sides by the coefficient of X which is 15
x= 11/15
PLEASE HELP I WILL GIVE BRAINLIEST
Step-by-step explanation:
A natural number is a positive whole number.
A whole number is a positive number with no fractions or decimals.
A interger is a whole number negative or positive.
A rational number is a number that terminates or continue with repeating digits.
A irrational number is a number that doesn't terminate or continue with repeating digits.
1. Rational Number
2. Natural,Whole,Interger,Rational
3. Whole,Rational,Interger
4. Rational
5.Irrational
6.Rational
7.Natural,Whole,Interger,Rational
8.Interger,Rational
9.Irrational
How do we derive the sum rule in differentiation? (ie. (u+v)' = u' + v')
It follows from the definition of the derivative and basic properties of arithmetic. Let f(x) and g(x) be functions. Their derivatives, if the following limits exist, are
[tex]\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h\text{ and }g'(x)\lim_{h\to0}\frac{g(x+h)-g(x)}h[/tex]
The derivative of f(x) + g(x) is then
[tex]\displaystyle \big(f(x)+g(x)\big)' = \lim_{h\to0}\big(f(x)+g(x)\big) \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{\big(f(x+h)+g(x+h)\big)-\big(f(x)+g(x)\big)}h \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{\big(f(x+h)-f(x)\big)+\big(g(x+h)-g(x)\big)}h \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{f(x+h)-f(x)}h+\lim_{h\to0}\frac{g(x+h)-g(x)}h \\\\ \big(f(x)+g(x)\big)' = f'(x) + g'(x)[/tex]
If LM = 9x + 27 and RS = 135, find x.
Answer:
x=12
Step-by-step explanation:
LM = RS
9x+27 = 135
Subtract 27 from each side
9x+27-27 =135-27
9x=108
Divide each side by 9
9x/9 = 108/9
x = 12
the boxes are equivalent so the one with a single dash is equal to the other with a single dash.
the one with 2 dashes is equal to the other with 2 dashes so on and so forth
SR=LM
LM=9x+27
RS=135
9x+27=135
so I solve it in my own weird way but you can solve it differently. 135-27=108
108/9=12
so your answer is 12
Where did term “infinity” come from
Solve for x in the triangle. Round your answer to the nearest tenth.
Answer:
x = 6.2
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan theta = opp / adj
tan 32 = x/ 10
10 tan 32 = x
x=6.24869
Rounding to the nearest tenth
x = 6.2
Answer:
x=6.2 (Rounded to the nearest tenth)
Step-by-step explanation:
This problem gives you an angle(32°), and ask for the dimension of the opposite side to that angle(x), along with another dimension the adjacent side(10).
Since you have the opposite and adjacent sides, you can use tangent. opposite (x) over adjacent (10). Tan(32) =[tex]\frac{x}{10}[/tex]. You want (x) so multiply tan(32) by 10. Then round to the nearest tenth.
Remember to put calculator in degree mode!
tan (32) = 0.6248693519 multiply by ten 6.248693519. Round to nearest 10th 6.2.
Hope this helps!
If a line has a midpoint at (2,5), and the endpoints are (0,0) and (4,y), what is the value of y? Please explain each step for a better understanding:)
Answer:
y = 10
Step-by-step explanation:
To find the y coordinate of the midpoint, take the y coordinates of the endpoints and average
(0+y)/2 = 5
Multiply each die by 2
0+y = 10
y = 10
Julie assembles shelves for a department store and gets paid $3.25 per shelf. She can assemble 5 per hour and works 8 hours per day. Determine Julie’s gross pay for 1 week
Pay per shelf = $3.25
No of shelfs per hour = 5
Total hours per day = 8
Total days to find pay of = 7
= 3.25×5×8×7
= 910
Therefore she is paid $910 after 1 week.
Must click thanks and mark brainliest
Classify the triangle as acute, right, or obtuse and as equilateral, isosceles, or scalene.
9514 1404 393
Answer:
(d) Right, scalene
Step-by-step explanation:
The little square in the upper left corner tells you that is a right angle. Any triangle with a right angle is a right triangle. This one is scalene, because the sides are all different lengths.
__
Additional comment
An obtuse triangle cannot be equilateral, and vice versa.
An equilateral triangle has all sides the same length, and all angles the same measure: 60°. It is an acute triangle.
Please help I’ll mark as brainlist
Answer:
Ekta and Preyal
Step-by-step explanation:
Two lateral faces of a rectangular pyramid have a base length of 10 inches and a height of 15 inches. The other two lateral faces have a base length of 18 inches and a height of 13 inches. What is the surface area of the rectangular pyramid?
The surface area of the rectangular pyramid is the sum of the area of its individual surface. The surface area of the pyramid is 384 square inches.
Each surface of the rectangular pyramid has the shape of a triangle.
So that the area of each surface of the pyramid = [tex]\frac{1}{2}[/tex] x base x height.
From the given question;
The area of one of the two lateral faces = [tex]\frac{1}{2}[/tex] x b x h
= [tex]\frac{1}{2}[/tex] x 10 x 15
The area of one of the two lateral faces = 75 square inches
Thus,
the area of the first two given lateral faces = 2 x 75
= 150 square inches
The area of the first two given lateral faces = 150 square inches
Also,
The area of one of the other two lateral faces = [tex]\frac{1}{2}[/tex] x b x h
= [tex]\frac{1}{2}[/tex] x 18 x 13
The area of one of the other two lateral faces = 117 square inches
So that;
the area of the first two other lateral faces = 2 x 117
= 234 square inches
The area of the first two other lateral faces = 234 square inches
Thus,
the surface area of the pyramid = 150 + 234
= 384 square inches
Therefore, the surface area of the rectangular pyramid is 384 square inches.
For more clarifications: https://brainly.com/question/23564399
Answer:
Step-by-step explanation:
The surface area of the rectangular pyramid is the sum of the area of its individual surface. The surface area of the pyramid is 384 square inches.
Each surface of the rectangular pyramid has the shape of a triangle.
So that the area of each surface of the pyramid = x base x height.
From the given question;
The area of one of the two lateral faces = x b x h
= x 10 x 15
The area of one of the two lateral faces = 75 square inches
Thus,
the area of the first two given lateral faces = 2 x 75
= 150 square inches
The area of the first two given lateral faces = 150 square inches
Also,
The area of one of the other two lateral faces = x b x h
= x 18 x 13
The area of one of the other two lateral faces = 117 square inches
So that;
the area of the first two other lateral faces = 2 x 117
= 234 square inches
The area of the first two other lateral faces = 234 square inches
The area of the rectangle is B x H = 18 x 10 = 180 square inches
Thus,
the surface area of the pyramid = 150 + 234+180
= 564 square inches
Therefore, the surface area of the rectangular pyramid is 564 square inches.
result of 5 and 75 with dividid by 3
Answer:
your answer is 30
Step-by-step explanation:
I hope this help
Help ASAP please :))
Image attached
help help help help
Answer:
abc is a triangle so ,
a is ( 9,6 )
b is ( 9,3 )
and c is ( 3,3 )