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ololo11 [35]
3 years ago
12

What are the solutions to the following system of equations?

Mathematics
2 answers:
Goryan [66]3 years ago
5 0

the answer to this question is (3,11) and (-3,-7)

Sholpan [36]3 years ago
3 0
We will express second equation in first.

y = 3x + 2
after that we get:

3x + 2 = x^2 + 3x - 7
x^2 -9 = 0
x^2 = 9
x1 = 3
x2 = -3

y1 = 3*3+2 = 11
y2 = -3*3 + 2 = -7

we get pairs:
(3,11) and (-3,-7)

Answer is first option.
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Anuta_ua [19.1K]

Answer:

1/14

Step-by-step explanation:

There is only 1 common multiple of 4 and 6 between 1 and 14.

So the probability is:

P = \frac{1}{14}

8 0
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Where is the function increasing? Please help, due soon
kaheart [24]
(Negative Infiniti, -1)
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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
Which of the following are identities? Check all that apply
Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}

Double angle tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula sin2\alpha =2sin\alpha *cos\alpha

sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)

D) False No further development towards a Trig Identity

[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)

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Hey can you please help me posted picture of question
shepuryov [24]
The correct answer is option C.

The solution is shown below:

(8 - 3i)(6 + 5i)

= 8(6 +5i) - 3i(6 + 5i)

= 48 + 40i - 18i - 15i²

= 22i + 48 - 15(-1)

= 22i + 48 + 15

= 22i + 63

= 63 + 22i
6 0
3 years ago
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