Answer:
She's "Hot"!
Step-by-step explanation:
Find all possible values of α+
β+γ when tanα+tanβ+tanγ = tanαtanβtanγ (-π/2<α<π/2 , -π/2<β<π/2 , -π/2<γ<π/2)
Show your work too. Thank you!
Answer:
[tex]\rm\displaystyle 0,\pm\pi [/tex]
Step-by-step explanation:
please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation
===========================
we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first
cancel tanγ from both sides which yields:
[tex] \rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) [/tex]
factor out tanγ:
[tex]\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)[/tex]
divide both sides by tanαtanβ-1 and that yields:
[tex]\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}[/tex]
multiply both numerator and denominator by-1 which yields:
[tex]\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)[/tex]
recall angle sum indentity of tan:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta ) [/tex]
let α+β be t and transform:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( t) [/tex]
remember that tan(t)=tan(t±kπ) so
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi ) [/tex]
therefore when k is 1 we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi ) [/tex]
remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal which yields:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm \pi [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }[/tex]
when is 0:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 ) [/tex]
likewise by Opposite Angle Identity we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal therefore:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm 0 [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }[/tex]
and we're done!
Answer:
-π, 0, and π
Step-by-step explanation:
You can solve for tan y :
tan y (tan a + tan B - 1) = tan a + tan y
Assuming tan a + tan B ≠ 1, we obtain
[tex]tan/y/=-\frac{tan/a/+tan/B/}{1-tan/a/tan/B/} =-tan(a+B)[/tex]
which implies that
y = -a - B + kπ
for some integer k. Thus
a + B + y = kπ
With the stated limitations, we can only have k = 0, k = 1 or k = -1. All cases are possible: we get k = 0 for a = B = y = 0; we get k = 1 when a, B, y are the angles of an acute triangle; and k = - 1 by taking the negatives of the previous cases.
It remains to analyze the case when "tan "a" tan B = 1, which is the same as saying that tan B = cot a = tan(π/2 - a), so
[tex]B=\frac{\pi }{2} - a + k\pi[/tex]
but with the given limitation we must have k = 0, because 0 < π/2 - a < π.
On the other hand we also need "tan "a" + tan B = 0, so B = - a + kπ, but again
k = 0, so we obtain
[tex]\frac{\pi }{2} - a=-a[/tex]
a contradiction.
Question 4 Multiple Choice Worth 4 points)
(01.02 LC)
What is the solution for the equation 6x - 8 = 4x?
Answer:
Algebra
Step-by-step explanation:
(+) it's the same thing btw6x -8 = 4x
(collect like terms) meaning the numbers with x go over the " = " sign
making it = -8 = 4x -6x
the signs change when it crosses over so it becomes that
-8 = -2x
-8 ÷ -2 = 4 (cause - ÷ by - is + )
What is another name for CD?
Answer:
I think album is another name of CD.
Which of the following is the function for the graph shown?
Answer:
D
Step-by-step explanation:
The zeros from the graph, where it crosses the x- axis are
x = - 2 and x = 3 , then the corresponding factors are
(x + 2) and (x - 3) , then
y = a(x + 2)(x - 3) ( where a is a multiplier )
To find a substitute any point on the graph into the equation
Using (0, - 6 )
- 6 = a(0 + 2)(0 - 3) = a(2)(- 3) = - 6a ( divide both sides by - 6 )
1 = a
y = (x + 2)(x - 3) ← expand using FOIL
y = x² - x - 6 → D
Mrs. Nygaard needs 12 hours to grade all of her students’ projects. She made a chart to show how much time she could spend grading the projects during the week. Project Grading Day Hours Grading Monday 1 and three-fourths Tuesday 1 and one-half Wednesday 1 and one-fifth Thursday 2 Friday 1 and one-fourth How many hours will Mrs. Nygaard need to work over the weekend to finish grading the projects? 1 and three-fifths hours 4 and StartFraction 3 over 10 EndFraction hours 6 and one-half hours 19 and StartFraction 7 over 10 EndFraction hours
Answer:
B
Step-by-step explanation:
Option B is correct, 4 and 3/10 hours will Mrs. Nygaard need to work over the weekend to finish grading the projects.
What is Fraction?A fraction represents a part of a whole.
To find the total amount of time Mrs. Nygaard has available during the week to grade projects, we need to add up the hours for each day:
1 and three-fourths + 1 and one-half + 1 and one-fifth + 2 + 1 and one-fourth = 7 and five-tenths hours
This means that Mrs. Nygaard has 7.5 hours during the week to grade projects.
If she needs 12 hours to grade all the projects, then she will need to work for an additional:
12 - 7.5 = 4.5 hours over the weekend to finish grading the projects.
Therefore, 4 and 3/10 hours required to Mrs. Nygaard need to work over the weekend to finish grading the projects.
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Oof, someone please help asap! I don't recall ever seeing this type of question before!
Answer:
90
Step-by-step explanation:
You are looking for the highest common factor for 270 and 360.
Factor the 2 numbers
270 = 2 * 5 * 3 * 3 * 3
360 = 2 * 2 * 2 * 3 * 3 * 5
Each of the numbers has a 5
Each of the numbers has two threes
Each of the numbers has one 2
So the answer is 2 * 3*3 * 5
[tex] \frac{x}{2} + \frac{6}{x } = 4[/tex]
using quadratic equation....help me if you can
Find the x in the kite below
Answer:
x = 5
Step-by-step explanation:
comment if you need explanation
Answer:
Step-by-step explanation:
It just so happens that x is the hypotenuse in the right triangle with sides 3 and 4. To find x we use Pythagorean's Theorem:
[tex]x^2=3^2+4^2[/tex] and
[tex]x^2=9+16[/tex] and
[tex]x^2=25[/tex] so
x = 5
Which expression is equivalent to 5g+9+8g+4?
[tex]\huge\textsf{Hey there!}[/tex]
[tex]\large\textsf{5g + 9 + 8g + 4}\\\\\huge\textsf{COMBINE the LIKE TERMS}\\\\\large\textsf{5g + 8g + 9 + 4}\\\\\large\textsf{5g + 8g = \bf 13g}\\\\\large\textsf{9 + 4 = \bf 13}\\\\\boxed{\huge\textsf{= 13g + 13}}\large\checkmark\\\\\\\\\\\\\boxed{\boxed{\huge\textsf{Answer: \bf 13g + 13}}}\huge\checkmark\\\\\\\\\\\\\\\\\large\textsf{Good luck on your assignment and enjoy your day!}\\\\\\\\\\\frak{Amphitrite1040:)}[/tex]
???????????????????????????
Answer: its 20 I think
Answer:
x = 50
I hope this help the side note also help me a lot as well
Help please guys if you don’t mind
Step-by-step explanation:
Step 2:
[tex] ({15x}^{2} - 9x) + (5x - 3)[/tex]
Stepc3:
[tex]3x(5x - 3)[/tex]
Evaluate the integral.
Answer:
D
Step-by-step explanation:
[tex]\int\limits^4_1 {\frac{7^{lnx}}{x} } \, dx \\put~ln~x=y\\diff.\\\frac{1}{x} dx=dy\\when~x=1,y=ln1=0\\when x=4,y=ln~4\\\int\limits^{ln4}_0 {7^y} \, dy\\=\frac{7^y}{ln7} |0 ~\rightarrow~ ln 4\\= \frac{7^{ln4}}{7}[/tex]
The area of a square garden is 40000/1600 square meters. Find the side length of the garden in meters.
Answer:
5 ..... 144/16 = 25 sqrt(25) = 5
Step-by-step explanation:
Which is the best estimate of 162% of 79?
I think it's answer is 79 just guess
8,X,20 are in arithmetic progression,find the value of "x".
Answer:
x = 14
Step-by-step explanation:
Since the terns form an arithmetic progression then they have a common difference d , that is
a₂ - a₁ = a₃ - a₂
x - 8 = 20 - x ( add x to both sides )
2x - 8 = 20 ( add 8 to both sides )
2x = 28 ( divide both sides by 2 )
x = 14
Help please-- Given circle O below, if arc GH and arc HJ are congruent, what is the measure of chord line HJ?
Answer:
the answer is D
Step-by-step explanation:
Find the distance between each pair of points. Round to the nearest tenth if necessary.
(4,2) and (-6, -6)
Answer:
Radical (20)
Step-by-step explanation:
Radical ( (4-6)² + (2-6)²)) =radical ( 4+16) = radical (20)
Nadia is mountain climbing. She started at an altitude of 19.26 feet below sea level and then changed her altitude by climbing a total of 5,437.8 feet up from her initial position. What was Nadia’s altitude at the end of her climb?
Answer:
5418.54 ft
Step-by-step explanation:
So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.
-19. 26+ 5437.8= 5418.54
Now just add on the units!
Hope this helps!
Answer:
Answer:
5418.54 ft
Step-by-step explanation:
Answer:
5418.54 ft
Step-by-step explanation:
So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.
-19. 26+ 5437.8= 5418.54
Find the missing side. Round to the nearest tenth. (ITS DUE IN THE MORNING PLEASE HELP)
24.
A) 14.2
C) 13.8
B) 9.2
D) 15.7.
25.
A) 37.6
B)30.8
C) 45.1
D)5.5
Answer:
24. A)14.2
25. D)5.5
Step-by-step explanation:
I hope it help for you keep safe
Rule 1: Multiply by 2 then add 1 starting from 1.
Rule 2: Divide by 2 then add 4 starting from 40.
Sequence 1
Sequence 2
Ordered Pairs
PLZ ANSWER FOR Brainiest
Answer:
1×2+1=3
3×2+1=7
7×2+1=15
15x2+1=31
31×2+1=63
40÷2+4=24
24÷2+4÷16
16÷2+4=12
12÷2+4=10
10÷2+4=9
Lines L and M are parallel.
Answer:
52°
Step-by-step explanation:
The angle 38° and m∠1 equal 90°, so 90-38=52.
Answer:
thats is an obtuse so if 2 is 38 then 1 should be 120 but u add those together u get 158 so it shold be 120 but if not then try looking explamles up
Step-by-step explanation:
please help me! im having trouble
Answers:
3 chocolate cones
1 strawberry cones
===========================================================
Explanation:
Let's isolate the variable c in the first equation
c+s = 4
c = 4-s
we subtract s from both sides to get c all by itself. This will then be plugged into the second equation. I'm using the substitution property.
1.75c + 1.3s = 6.55
1.75(4-s) + 1.3s = 6.55 ..... replace c with 4-s
1.75*4 + 1.75*(-s) + 1.3s = 6.55
7 - 1.75s + 1.3s = 6.55
-0.45s + 7 = 6.55
-0.45s = 6.55 - 7
-0.45s = -0.45
s = -0.45/(-0.45)
s = 1
So he bought 1 strawberry cone. Use this value of s to find c
c = 4-s
c = 4-1
c = 3
This says he also bought 3 chocolate cones.
-------------------
As a check, we see that c+s = 3+1 = 4, showing that he bought 4 cones total.
Also,
1.75c + 1.3s = 1.75*3 + 1.3*1 = 6.55
indicating he spent $6.55 total for the four cones. This matches with what the instructions tell us, so the answer is confirmed.
Answer:
Step-by-step explanation:
[tex]\left \{ {{c+s=4} \atop {1.75c+1.3s=6.55}} \right.[/tex]
1.75c + 1.75s = 1.75 × 4 ........ (1)
1.75c + 1.3s = 6.55 ........ (2)
(1) - (2)
0.45s = 0.45 ⇒ s = 1
c + 1 = 4 ⇒ c = 3
[ 3 ] chocolate [ 1 ] strawberry
1. Paul uses a coordinate plane to design
his model town layout.
Paul moves the market 2 units left and 3
units down. He says the ordered pair for
the new location of the market is (0,6).
Explain Paul's mistake and write the
correct ordered pair for the new location of
the market.
PLZ ALSO INCLUDE WHAT HIS MISTAKE WAS!
ANSWER FOR Brainiest!!!
The graph of y=x^3+x^2-6x is shown....
hello,
" a turning point is defined as the point where a graph changes from either increasing to decreasing, or decreasing to increasing"
a)
[tex]y=x^3+x^2-6x\\\\y'=3x^2+2x-6=0\\x=\dfrac{-2-\sqrt{76} }{6} \approx{-1.786299647...}\\or\\x=\dfrac{-2+\sqrt{76} }{6} \approx{1.1196329...}\\[/tex]
b)
Zeros are -3,0,2.
Sol={-3,0,2}
The solution of the graph function y=x³+x²-6x are -3 , 0 and 2
What is graph?The link between lines and points is described by a graph, which is a mathematical description of a network. A graph is made up of certain points and the connecting lines. It doesn't matter how long the lines are or where the points are located. A node is the name for each element in a graph.
We have the function
y=x³+x²-6x
now, equating it to 0
x³+x²-6x = 0
x² + x - 6= 0
x² - 3x + 2x -6 =0
x(x -3) + 2(x -3)
x= 3 and -2
Now, ew can see from the that the equation is touching the x-axis at three points and it will represent three zeroes of the equation.
So, the solution of the graph are -3 , 0 and 2
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help or i will fail my acellus
Answer:
I think it's 155 cm
Step-by-step explanation:
=(5×5×3)+(5×2×4)
= 75+40
= 155 cm2
help! urgent asap!!!!!!!!!!
Answer:
A
Step-by-step explanation:
This is because the answer you picked means congruent to, however these two triangles are not the same size
The line y = 2x + 6 cuts the x-axis at A and the y-axis at B. Find
(a) the length of AB,
(b) the shortest distance of O to AB, where O is the origin (0,0)
Answer:
(a)
[tex]3 \sqrt{5} [/tex]
(b)
[tex] \frac{6}{ \sqrt{5} } [/tex]
Step-by-step explanation:
A(-3,0)
B(0,6)
[tex]d = \sqrt{{( - 3 - 0)}^{2} + {(0 - 6)}^{2} } = \sqrt{9 + 36} = 3 \sqrt{5} [/tex]
[tex]d = \frac{ax0 + by0 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } [/tex]
2x-y+6=0
a=2, b=-1, c=6
x0=0, y0=0
[tex]d = \frac{6}{ \sqrt{4 + 1} } = \frac{6}{ \sqrt{5} } [/tex]
find the slope of the line passing through the points (-2,5) and (3/2,2)
Answer:
slope = - [tex]\frac{6}{7}[/tex]
Step-by-step explanation:
Calculate the slope m using the slope formula
m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]
with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = ([tex]\frac{3}{2}[/tex], 2)
m = [tex]\frac{2-5}{\frac{3}{2}-(-2) }[/tex]
= [tex]\frac{-3}{\frac{3}{2}+2 }[/tex]
= [tex]\frac{-3}{\frac{7}{2} }[/tex]
= - 3 × [tex]\frac{2}{7}[/tex]
= - [tex]\frac{6}{7}[/tex]
which of the following are identities? check all that apply.
A. (sinx + cosx)^2= 1+sin2x
B. sin6x=2 sin3x cos3x
C. sin3x/sinxcosx = 4cosx - secx
D. sin3x-sinx/cos3x+cosx = tanx
Answer: (a), (b), (c), and (d)
Step-by-step explanation:
Check the options
[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]
[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]
[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]
[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]
Thus, all the identities are correct.
A. Not an identity
B. An identity
C. Not an identity
D. An identity
To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:
A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]
To check if this is an identity, let's expand the left-hand side (LHS):
[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]
Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:
[tex]\(\sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\)[/tex]
The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.
B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]
Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]
[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
The equation holds true with the double-angle identity, so option B is an identity.
C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]
To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.
Recall that [tex]\(\sec x = \frac{1}{\cos x}\):[/tex]
[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]
Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]
[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]
Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]
Now, the left-hand side (LHS) becomes:
[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]
Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]
[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]
So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.
D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]
To check if this is an identity, we can use the sum-to-product trigonometric identities:
[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]
Let A = 3x and B = x:
[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]
Now, we can rewrite the expression:
[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]
The equation holds true, so option D is an identity.
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The figure below shows the design of a truss to support the weight of a roof. In this truss design, ∠A⩭∠D and ĀF⩭DF.
Answer:
[tex]m\angle A= 30^{\circ}[/tex]
Step-by-step explanation:
In the question, we're given [tex]\angle A\cong \angle D[/tex]. Therefore, the measure of these two angles must be equal.
To find the value of [tex]x[/tex], set these two angles equal to each other:
[tex]4x-26=x+16[/tex]
Add 26 and subtract [tex]x[/tex] from both sides:
[tex]3x=42[/tex]
Divide both sides by 3:
[tex]x=\frac{42}{3}=14[/tex]
Since [tex]\angle A[/tex] was labelled as [tex]4x-26[/tex], substitute [tex]x=14[/tex] to find its measure:
[tex]\angle A=4(14)-26,\\\angle A=56-26,\\\angle A=\boxed{30^{\circ}}[/tex]
You can also substitute [tex]x=14[/tex] into the label of angle D as angle A is congruent to angle D for easier calculations ([tex]14+16=\boxed{30^{\circ}}[/tex]).