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Mademuasel [1]
3 years ago
5

Which properties are best used prove a quadrilateral is a square?

Mathematics
1 answer:
adell [148]3 years ago
8 0

Answer:

a, b, c, & d

Step-by-step explanation:

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Step-by-step explanation:

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How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?
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(5.8 x 10) - (2.2 x 105)
3241004551 [841]

The answer you're looking for is -173

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Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value
vodomira [7]

Answer:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1, \frac{\sqrt{3}}{2}  - 1

Step-by-step explanation:

we are given two <u>coincident</u><u> points</u>

\displaystyle  P( \sin(θ)+2,  \tan(θ)-2)   \: \text{and } \\  \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

since they are coincident points

\rm \displaystyle  P( \sin(θ)+2,  \tan(θ)-2)    = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

By order pair we obtain:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) =  \sin( \theta)   + 2 \\   \\  \displaystyle 3 \sin( \theta)  - 2  \cos( \theta)  + a =  \tan( \theta)  - 2\end{cases}

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using <u>substitution</u><u> method</u>

to do so, make a the subject of the equation.therefore from the second equation we acquire:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )=  \sin( \theta)   + 2 \\   \\  \boxed{\displaystyle  a =  \tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta) } \end{cases}

now substitute:

\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta)   \}=  \sin( \theta)   + 2

distribute:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  - 6 \sin( \theta) \cos( \theta)    + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

collect like terms:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)     + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

rearrange:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta)  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + =  \sin( \theta)   + 2

by <em>Pythagorean</em><em> theorem</em> we obtain:

\rm\displaystyle \displaystyle 4  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)   + 2

cancel 4 from both sides:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)    - 2

move right hand side expression to left hand side and change its sign:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2  =  0

factor out sin:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2  =  0

factor out 2:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1)  + 2(- 2\cos( \theta) + 1 ) =  0

group:

\rm\displaystyle \displaystyle ( \sin (θ)   + 2)(- 2 \cos(θ)+1)  =  0

factor out -1:

\rm\displaystyle \displaystyle -  ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

divide both sides by -1:

\rm\displaystyle \displaystyle   ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we acquire:

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)   + 2 = 0 \\ \displaystyle2 \cos(θ) - 1=  0 \end{cases}

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)     \neq  - 2 \\ \displaystyle2 \cos(θ) =  1\end{cases}

divide both sides by 2:

\rm\displaystyle \displaystyle \displaystyle \cos(θ) =   \frac{1}{2}

by unit circle we get:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

so when θ is 60° a is:

\rm \displaystyle a =  \tan(  {60}^{ \circ} )  - 2 - 3 \sin(  {60}^{ \circ} )   +  2  \cos(  {60}^{ \circ} )

recall unit circle:

\rm \displaystyle a =   \sqrt{3}  - 2 -  \frac{ 3\sqrt{3} }{2}   +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1

when θ is 300°

\rm \displaystyle a =  \tan(  {300}^{ \circ} )  - 2 - 3 \sin(  {300}^{ \circ} )   +  2  \cos(  {300}^{ \circ} )

remember unit circle:

\rm \displaystyle a =  -  \sqrt{3}   - 2  +   \frac{3\sqrt{ 3} }{2}  +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

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The answer will be 1.8
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