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stealth61 [152]
3 years ago
13

Which system of inequalities is graphed below

Mathematics
2 answers:
Anna [14]3 years ago
6 0
The closest to this was letter C

hope that helps, God bless! 
suter [353]3 years ago
4 0

Answer:

<em>Option D</em>

Step-by-step explanation:

Let's explain why the Option D is correct :

  • The Parabola equation:  $x \leq y^2 - 8$

This is a parabola equation <em>with vertex at x = -8.</em>

Despite that the "traditional equation" is $ y=x^2 + b $  in this case  the parabola equation is plotted in the X axis<em> and </em>has a $ sign which indicates that X will take values less and equal than $ y^2 -8 $

<u><em>The graph of the parabola is attached to the answer.</em></u>

  • The hyperbola equation: $ {x^2 /{3^2}} - {y^2 / {5^2}} > 1 $

This is a hyperbola equation with center at (0,0) and vertex's : V1=(3;0) and V2=(-3;0)

The two points (0;5) and (0;-5) on the Y axis are not on the hyperbola.

It's similar to the previous case of the parabola, Despite that the "traditional equation"  of the Hyperbola : $ {x^2 /{a^2}} - {y^2 / {b^2}} = 1 $  in this case  the hyperbola  equation has a $ > $ sign which indicates that hyperbola will take values greater than those who correspond to $ {x^2 /{3^2}} - {y^2 / {5^2}} = 1 $

<u><em>The graph of the Hyperbola is also attached</em></u>

Finally we deduce  that the  D system of inequalities is the correct one.

<u><em>Also the answer is attached</em></u>.

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Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

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3 years ago
Simplify -2X + 4Y -16
Novosadov [1.4K]
The answer is -2(x+6)
6 0
3 years ago
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TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?
egoroff_w [7]

Answer:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

Step-by-step explanation:

3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

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4 years ago
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Neko [114]

There are no solutions to the given system of equations.

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4 0
3 years ago
If Samantha can pay off her loan in 36 months at a 10% interest rate rather than in 48 months at a 12% interest rate, how much m
Ymorist [56]
Hi there
The simple interest formula is
I=prt
I interest changes
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R interest rate
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The interest in 36 months at a 10%
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The interest in 48 months at a 12%
6,000×0.12×(48÷12)=2,880
she will save
2,880−1,800=1,080

Good luck!
7 0
3 years ago
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