A distribution has the five-number summary shown below. What is the
interquartile range (IQR) of this distribution?
12, 21, 43, 62, 71

Answer:

[tex]\large\boxed{y=2}[/tex]

Step-by-step explanation:

In this question, we're going to solve the equation.

What we would do is plug in the given information in the right variable. In this case, we would plugin -8 to x, since x = -8

We are solving for y, so we would nee to find how much y = ?

Your equation should look like this:

[tex]4(-8)+9y=-14[/tex]

Lets solve:

[tex]4(-8)+9y=-14\\\\\text{Multiply}\, 4(-8)\\\\-32+9y=-14\\\\\text{Add 32 to both sides}\\\\9y=18\\\\\text{Divide}\\\\y=2[/tex]

When you're done solving, you should get y = 2

This means that y = 2

Answer:

[tex]\text{The point-slope equation of a line is:}[/tex]

[tex]C.\ y-y_0=m(x-x_0)[/tex]

[tex]m-\text{slope}\\\\(x_0,\ y_0)-\text{point on a line}[/tex]

Answer:

C. No because Angie is not representative of the population in the algebra class

Step-by-step explanation:

I just took the test and this was correct

No, this is not an accurate predictor

Since Angie is the student of an  advanced statistics course but she gives the algebra test and scored 80%. We know that the advanced statistics is the higher level of math so it is not an accurate predictor.

An accurate predictor should be applied when the person is 100% confirmed to the related thing. So in the given situation, the student is not 100% confirm.

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Answer: 35

Step-by-step explanation:

The measure of ∠Z is 35°

The correct answer is an option (D)

For given question,

We have been given a parallelogram WXYZ.

The measure of angle X is 35°

We know that, the opposite angles of parallelogram are equal.

⇒ ∠X = ∠Z

⇒ ∠Z = 35°

Therefore, the measure of ∠Z is 35°

The correct answer is an option (D)

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$9.38

0.0625×150=$9.38

Have a good day

Answer:

First angle = 50, second angle = 26 and the third angle = 104 degrees.

Largest angle is 104 degrees.

Step-by-step explanation:

Let the second angle be x degrees, the the first will be x + 24 and the third is

4x degrees.

x + x+ 24 + 4x = 180

6x = 180 - 24

6x = 156

x = 26

so the other 2 angles are 26 + 24 = 50 and 4*26 = 104 degrees.  

Triangle is the polygon with three edges, three vertices and three angle. The total number of degrees in a triangle is always 180 degrees. The value of the largest angle in the given triangle is 104 degrees.

Let The measure of the second angle is x degrees.

The measure of the first angle is 24 degrees more than the measure of the second angle. Thus first angle is (24+x) degrees.

The third angle is four times the measure of the second angle. Thus the measure of the third angle is 4x.

Triangle is the polygon with three edges, three vertices and three angle. The total number of degrees in a triangle is always 180 degrees.

As we know that the sum of the all the three angle in the triangle is 180 degrees. Therefore,

[tex](24+x)+x+4x=180[/tex]

[tex]6x=124-48[/tex]

[tex]6x=156[/tex]

[tex]x=\dfrac{156}{6}[/tex]

[tex]x=26[/tex]

Hence the the measure of the second angle is 26 degrees. Now The measure of the first angle is 24 degrees more than the measure of the second angle. Thus,

[tex]=24+26[/tex]

[tex]=50[/tex]

Hence measure of the first angle is 50 degrees. Now the third angle is four times the measure of the second angle. Thus,

[tex]=4\times 26=104[/tex]

Hence, the measure of the third angle is 104 degrees.

Thus the value of the largest angle in the given triangle is 104 degrees.

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Answer:

See below.

Step-by-step explanation:

You fill the 5 gallon up to the top then you pour 3 gallons into the 3 gallon jug, leaving 2 gallons in the  bigger jug.

You throw the contents of the 3 gallon jug away then transfer the 2 gallons from the bigger jug into the 3 gallon jug. So you have  2 gallons in the 3 gallon jug.

You then fill the 5 gallon jug and pour 1 gallon into the 3 gallon jug, so you are left with 4 gallons of water in the 5 gallon jug.

[tex]|\Omega|=6^2=36\\A=\{(3,5),(5,3),(4,4),(2,6),(6,2)\}\\|A|=5\\\\P(A)=\dfrac{5}{36}\approx13.9\%[/tex]

Answer:

[tex]\large\boxed{y=3x-11}[/tex]

Step-by-step explanation:

[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===================================\\\\y=3x-4\to m_1=3\\\\\text{Therefore}\ m_2=3.\\\\\text{We have the equation:}\ y=3x+b.\\\\\text{Put the coorsdinates of the point (2, -5) to the equation of the line}\\\\-5=3(2)+b\\-5=6+b\qquad\text{subtract 6 from both sides}\\-11=b\to b=-11[/tex]

Answer:

[tex]\large\boxed{y=\dfrac{1}{4}x+4}[/tex]

Step-by-step explanation:

[tex]Let:\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\============================\\\\\text{We have the equation of the line:}\ y=-4x+9\to m_1=-4.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-4}=\dfrac{1}{4}.\\\\\text{Put the value of the slope and the coodrdinates of the given point (4, 5)}\\\text{to the equation of a line}\ y=mx+b:\\\\5=\dfrac{1}{4}(4)+b\\\\5=1+b\qquad\text{subtract 1 from both sides}\\\\4=b\to b=4\\\\\text{Finally:}\\\\y=\dfrac{1}{4}x+4[/tex]

tangent is less than 0 or tan(θ) < 0, is another way to say tan(θ) is negative, well, that only happens on the II Quadrant and IV Quadrant, where sine and cosine are different signs, so we know θ is on the II or IV Quadrant.

[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{3}}\qquad \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}[/tex]

[tex]\bf \pm\sqrt{3^2-2^2}=a\implies \pm\sqrt{5}=a\implies \stackrel{\textit{II Quadrant}}{-\sqrt{5}=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{5}}}{\stackrel{hypotenuse}{3}}~\hfill[/tex]

Answer:

5

Step-by-step explanation:

[tex]A_{s} = l^2[/tex], where l - is a side

[tex]l = \sqrt{A_s}[/tex]

[tex]l = \sqrt{25} = 5[/tex]

Answer:

Step-by-step explanation:

Look at the picture.

9. The perimeter of the rectangle l × w:

P = 2l + 2w

Substitute l = 2x + 8 and w = x + 8:

P = 2(2x + 8) + 2(x + 8)     use the distributive property

P = (2)(2x) + (2)(8) + (2)(x) + (2)(8)

P = 4x + 16 + 2x + 16     combine like terms

P = (4x + 2x) + (16 + 16)

P = 6x + 32

10. Solve the equation:

6x + 32 = 158           subtract 32 from both sides

6x = 126     divide both sides by 6

x = 21

Put the value of x to the expression 2x + 8:

2(21) + 8 = 42 + 8 = 50

Answer:

Probably of getting both = 3/10 *100 = 30%

Step-by-step explanation:

According to the given statement we have four possible combinations:

Blue pants/white shirt, tan pants/white shirt, blue pants/blue shirt, blue pants/white shirt

.probability of getting tan pants = 2/4

probability of getting white shirt = 3/5

probably of getting both = 2/4 x 3/5 = 6/20 = 3/10

probably of getting both = 3/10 *100 = 30% ...

Answer:

Choice A

Step-by-step explanation:

Use the F. O. L. D method while solving

Answer:

there are 4 pints

Step-by-step explanation:

64 / 16 = 4

64 fluid ounces of lemonade equals to 4 pints. This is done by dividing the total fluid ounces by the number of fluid ounces in one pint.

The question is asking to convert 64 fluid ounces of lemonade into pints. To do this, you should know that 1 pint is equal to 16 fluid ounces. So, you can divide the total fluid ounces by the number of fluid ounces in one pint. In this situation, divide 64 by 16, which equals 4 pints. Therefore, 64 fluid ounces of lemonade equals to 4 pints.

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Answer:

see explanation

Step-by-step explanation:

Using Pythagoras' identity

The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other 2 sides.

Consider the right triangle on the right and calculate n

n² + 2² = 3²

n² + 4 = 9 ( subtract 4 from both sides )

n² = 5 ( take the square root of both sides )

n = [tex]\sqrt{5}[/tex]

If the triangle on the left is right then the square of the longest side must equal the sum of the squares on the other 2 sides.

The longest side is n = [tex]\sqrt{5}[/tex] ⇒ n² = ([tex]\sqrt{5}[/tex] )² = 5 and

([tex]\sqrt{2}[/tex] )² + ([tex]\sqrt{3}[/tex] )² = 2 + 3 = 5

Hence D is a right angle

For this case we must find an expression equivalent to:

[tex]\sqrt [7] {x} * \sqrt [7] {x} * \sqrt [7] {x}[/tex]

By definition of properties of powers and roots we have:

[tex]\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}[/tex]

So, rewriting the given expression we have:

[tex]x ^ {\frac {1} {7}} * x ^ {\frac {1} {7}} * x ^ {\frac {1} {7}} =[/tex]

To multiply powers of the same base we put the same base and add the exponents:

[tex]x ^ {\frac {1} {7} + \frac {1} {7} + \frac {1} {7}} =\\x ^ {\frac {3} {7}}[/tex]

Answer:

Option 1

Answer: Option 1.

Step-by-step explanation:

We need to remember that:

[tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}[/tex]

Then, having the expression:

[tex]\sqrt[7]{x}*\sqrt[7]{x} *\sqrt[7]{x}[/tex]

We can rewrite it in this form:

[tex]=x^{\frac{1}{7}}*x^{\frac{1}{7}} *x^{\frac{1}{7}}[/tex]

According to the Product of powers property:

[tex](a^m)(a^n)=a^{(m+n)}[/tex]

Then, the simplied form of the given expression, is:

[tex]=x^{(\frac{1}{7}+\frac{1}{7}+\frac{1}{7})}=x^\frac{3}{7}[/tex]

Answer:

The function f(x)=3x-10 has the greater value at x=8.

Step-by-step explanation:

In the table, x=8 corresponds to y=4.

In the function f(x)=3x-10 when x=8, y corresponds to f(8)=3(8)-10=24-10=14.

SInce 14>4 , the the function f(x)=3x-10 has a greater value at x=8 than the table does.

Answer:

x=y2/8+y/2+9/2

Step-by-step explanation:

Given:

directrix: x=2

focus = (6,-2)

Standard equation of parabola is given by:

(y - k)2 = 4p (x - h)

where

directrix : x=h-p

focus=(h + p, k)

Now comparing the give value with above:

(h + p, k)= (6,-2)

k=-2

h+p=6

h=6-p

Also

directrix: x=h-p

h-p=2

Putting value of h=6-p in above

6-p-p=2

6-2p=2

-2p=2-6

-2p=-4

p=-4/-2

p=2

Putting p=2 in h-p=2

h=2+p

h=2+2

h=4

Putting k=-2, p=2, h=4 in standard equation of parabola we get:

(y - k)2 = 4p (x - h)

(y-(-2))^2 = 4(2) (x - 4)

(y+2)^2 = 8 (x - 4)

y2+4y+4=8x-32

y2+4y+4+32=8x

x=y2/8+4y/8+36/8

x=y2/8+y/2+9/2!

Answer:

Option C

Step-by-step explanation:

As per the contract:

"Any rental not received by midnight on the day it is due is subject to a late charge of $1.50 for each day it is late."

this makes option a: "Pay a $1.50 late fee per movie per day past the due date" responsibility of customer

"All rentals are due back by midnight of the due date as printed on the transaction receipt."

this makes option b: "Get his or her rentals back by midnight the day they are due." responsibility of customer

"Any rental not returned by the fifth day after the due date will be transferred to a sale. The Customer will then be required to pay the purchase price of the item in addition to five (5) days of late fees. "

this makes option d: "Pay the purchase price and five days of late fees for any item kept more than five days past the due date." responsibility of customer

Hence option C is correct: Return a rental item kept out thirteen days past the due date!

The option that is not the responsibility of the customer is Return a rental item kept out thirteen days past the due date.

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Answer:

6/11

Step-by-step explanation:

3 white circles, 3 black circles, so six circles out of the total 11 peices

Answer:

f(4) = 14

Step-by-step explanation:

f(x) = 3x+2

Let x=4

f(4) = 3(4)+2

    = 12+2

    = 14

Answer:

(x - 3)(x - 12)

Step-by-step explanation:

Given

x² - 15x + 36

Consider the factors of the constant term (+ 36) which sum to give the coefficient of the x- term (- 15)

The factors are - 3 and - 12, since

- 3 × - 12 = + 36 and - 3 - 12 = - 15, hence

x² - 15x + 36 = (x - 3)(x - 12) ← in factored form

The factored form of the polynomial x² − 15x + 36 is (x − 3)(x − 12). This can be obtained by using the steps of splitting the middle term method.

Given the polynomial, x² − 15x + 36

(-3x)×(-12x)=36x² and (-3x)+(-12x)= - 15x

-3x and -12x are the required factors of 36x²

x² - 3x - 12x+36

x² - 3x - 12x+36 = x(x - 3) - 12(x - 3)

                          =(x - 3)(x - 12)

Hence the factored form of the polynomial x² − 15x + 36 is (x − 3)(x − 12).

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[tex]\dfrac{d}{dx}(\dfrac{\sin(3x)}{x})[/tex]

First we must apply the Quotient rule that states,

[tex](\dfrac{f}{g})'=\dfrac{f'g-g'f}{g^2}[/tex]

This means that our derivative becomes,

[tex]\dfrac{\dfrac{d}{dx}(\sin(3x))x-\dfrac{d}{dx}(x)\sin(3x)}{x^2}[/tex]

Now we need to calculate [tex]\dfrac{d}{dx}(\sin(3x))[/tex] and [tex]\dfrac{d}{dx}(x)[/tex]

[tex]\dfrac{d}{dx}(\sin(3x))=\cos(3x)\cdot3[/tex]

[tex]\dfrac{d}{dx}(x)=1[/tex]

From here the new equation looks like,

[tex]\dfrac{3x\cos(3x)-\sin(3x)}{x^2}[/tex]

And that is the final result.

Hope this helps.

r3t40

Answer:

[tex]\frac{3\cos(3x)}{x}-\frac{\sin(3x)}{x^2}[/tex]

Step-by-step explanation:

If [tex]f(x)=\frac{\sin(3x)}{x}[/tex], then  

[tex]f(x+h)=\frac{\sin(3(x+h)}{x+h}=\frac{\sin(3x+3h)}{x+h}[/tex].

To find this all I did was replace old input, x, with new input, x+h.

Now we will need this for our definition of derivative which is:

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

Before we go there I want to expand [tex]sin(3x+3h)[/tex] using the sum identity for sine:

[tex]\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)[/tex]

[tex]\sin(3x+3h)=\sin(3x)\cos(3h)+\cos(3x)\sin(3h)[/tex]

So we could write f(x+h) as:

[tex]f(x+h)=\frac{\sin(3x)\cos(3h)+\cos(3x)\sin(3h)}{x+h}[/tex].

There are some important trigonometric limits we might need before proceeding with the definition for derivative:

[tex]\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1[/tex]

[tex]\lim_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0[/tex]

Now let's go to the definition:

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{\frac{\sin(3x)\cos(3h)+\cos(3x)\sin(3h)}{x+h}-\frac{\sin(3x)}{x}}{h}[/tex]

I'm going to clear the mini-fractions by multiplying top and bottom by a common multiple of the denominators which is x(x+h).

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{x(\sin(3x)\cos(3h)+\cos(3x)\sin(3h))-(x+h)\sin(3x)}{x(x+h)h}[/tex]

I'm going to distribute:

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)\cos(3h)+x\cos(3x)\sin(3h)-x\sin(3x)-h\sin(3x)}{x(x+h)h}[/tex]

Now I’m going to group xsin(3x)cos(3h) with –xsin(3x) because I see when I factor this I might be able to use the second trigonometric limit I mentioned.  That is xsin(3x)cos(3h)-xsin(3x) can be factored as xsin(3x)[cos(3h)-1].

Now the limit I mentioned:

[tex]\lim_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0[/tex]

If I let u=3h then we have:

[tex]\lim_{3h \rightarrow 0}\frac{\cos(3h)-1}{3h}=0[/tex]

If 3h goes to 0, then h goes to 0:

[tex]\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{3h}=0[/tex]

If I multiply both sides by 3 I get:

[tex]\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{h}=0[/tex]

I’m going to apply this definition after I break my limit using the factored form I mentioned for those two terms:

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)\cos(3h)-x\sin(3x)+x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}[/tex]

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)(\cos(3h)-1)+x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}[/tex]

[tex]f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)(\cos(3h)-1)}{x(x+h)h}+\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}[/tex]

So the first limit I’m going to write as a product of limits so I can apply the limit I have above:

[tex]f’(x)=\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{h} \cdot \lim_{h \rightarrow 0}\frac{x\sin(3x)}{x(x+h)}+\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}[/tex]

The first limit in that product of limits goes to 0 using our limit from above.

The second limit goes to sin(3x)/(x+h) which goes to sin(3x)/x since h goes to 0.

Since both limits exist we are good to proceed with that product.

Let’s look at the second limit given the first limit is 0. This is what we are left with looking at:

[tex]f’(x)=\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}[/tex]

I’m going to write this as a sum of limits:

[tex]\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)}{x(x+h)h}+\lim_{h \rightarrow 0}\frac{-h\sin(3x)}{x(x+h)h}[/tex]

I can cancel out a factor of x in the first limit.  

I can cancel out a factor of h in the second limit.

[tex]\lim_{h \rightarrow 0}\frac{\cos(3x)\sin(3h)}{(x+h)h}+\lim_{h \rightarrow 0}\frac{-\sin(3x)}{x(x+h)}[/tex]

Now I can almost use sin(u)/u goes to 1 as u goes to 0 for that first limit after writing it as a product of limits.  

The second limit I can go ahead and replace h with 0 since it won’t be over 0.

So this is what we are going to have after writing the first limit as a product of limits and applying h=0 to the second limit:

[tex]\lim_{h \rightarrow 0}\frac{\sin(3h)}{h} \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x+0)}[/tex]

Now the first limit in the product I’m going to multiply it by 3/3 so I can apply my limit as sin(u)/u->1 then u goes to 0:

[tex]\lim_{h \rightarrow 0}3\frac{\sin(3h)}{3h} \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x)}[/tex]

[tex]3(1) \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x)}[/tex]

So we can plug in 0 for that last limit; the result will exist because we do not have over 0 when replacing h with 0.

[tex]3(1)\frac{\cos(3x)}{x}+\frac{-\sin(3x)}{x^2}[/tex]

[tex]\frac{3\cos(3x)}{x}-\frac{\sin(3x)}{x^2}[/tex]

The given system of equation can be solved if x = 2/3 and y = -4/3.

So we need to find another system which can be solved by these values.

From the given 4 options lets find out which can be solved by x = 3/2 and y = -4/3

1.

y = -2y^2 -2 , x = y - 2

lets put the value of y here

y = -2(16/9)-2 = -32/9-2= -50/ 9

So not an equivalent equation.

2.

y = -2y^2 + 2 , x = y + 2

y = -2(16/9)+2=-32/9+2=-14/9

So not an equivalent system.

3.

y = -2y^2 - 8y - 8 , x = y + 2

y = -2(16/9)-8(-4/3)-8 = -32/9+32/3-8=64/9-8=-8/9

So not an equivalent system.

4.

y = -2y^2 + 8y - 8 , x = y -2

y = -2(16/9)+8(-4/3)-8=-32/9-32/3-8=-128/9-8=-22.22

So this is not an equivalent equation too

Answer:

C. [tex]\left \{ {{y=-2y^{2}-8y-8 } \atop {x=y+2}} \right.[/tex]

Step-by-step explanation:

Edge 2020

Answer:

Step-by-step explanation:

(f - g)(x) = f(x) - g(x)

f(x) = 3x + 2, g(x) = 2x - 2.

Substitute:

(f - g)(x) = (3x + 2) - (2x - 2) = 3x + 2 - 2x - (-2) = 3x + 2 - 2x + 2

combine like terms

(f - g)(x) = (3x - 2x) + (2 + 2) = x + 4

See the attached picture:

[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\sec^2\theta+1=0[/tex]

Recall that [tex]\tan^2\theta+1=\sec^2\theta[/tex], so we can write everything in terms of [tex]\tan\theta[/tex]:

[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\tan^2\theta=0[/tex]

Let [tex]x=\tan\theta[/tex], so that

[tex]x^3-\dfrac1{x-1}-x^2=0[/tex]

With some rewriting we get

[tex]x^3-x^2-\dfrac1{x-1}=0[/tex]

[tex]x^2(x-1)-\dfrac1{x-1}=0[/tex]

[tex]\dfrac{x^2(x-1)^2-1}{x-1}=0[/tex]

Clearly we cannot have [tex]x=1[/tex], or [tex]\tan\theta=1[/tex].

The numerator determines when the expression on the left reduces to 0:

[tex]x^2(x-1)^2-1=0[/tex]

[tex]x^2(x-1)^2=1[/tex]

[tex]\sqrt{x^2(x-1)^2}=\sqrt1[/tex]

[tex]|x(x-1)|=1[/tex]

[tex]x(x-1)=1\text{ or }x(x-1)=-1[/tex]

Completing the square gives

[tex]x(x-1)=x^2-x=x^2-x+\dfrac14-\dfrac14=\left(x-\dfrac12\right)^2-\dfrac14[/tex]

so that

[tex]\left(x-\dfrac12\right)^2=\dfrac54\text{ or }\left(x-\dfrac12\right)^2=-\dfrac34[/tex]

The second equation gives no real-valued solutions because squaring any real number gives a positive real number. (I'm assuming we don't care about complex solutions.) So we're left with only

[tex]\left(x-\dfrac12\right)^2=\dfrac54[/tex]

[tex]\sqrt{\left(x-\dfrac12\right)^2}=\sqrt{\dfrac54}[/tex]

[tex]\left|x-\dfrac12\right|=\dfrac{\sqrt5}2[/tex]

which again gives two cases,

[tex]x-\dfrac12=\dfrac{\sqrt5}2\text{ or }x-\dfrac12=-\dfrac{\sqrt5}2[/tex]

[tex]x=\dfrac{1+\sqrt5}2\text{ or }x=\dfrac{1-\sqrt5}2[/tex]

Then when [tex]x=\tan\theta[/tex], we can find [tex]\cot\theta[/tex] by taking the reciprocal, so we get

[tex]\boxed{\cot\theta=\dfrac2{1+\sqrt5}\text{ or }\cot\theta=\dfrac2{1-\sqrt5}}[/tex]

Answer: A. -1

Step-by-step explanation:

Edg 2020

Answer:

[tex]\large\boxed{y=(x-3)^2-3}[/tex]

Step-by-step explanation:

The vertex form of a quadratic equation y = ax² + bx + c:

y = a(x - h)² + k

(h, k) - coordinates of a vertex

We have the equation y = x² - 6x + 6.

Convert to the vertex form:

[tex]y=x^2-2(x)(3)+6\\\\y=\underbrace{x^2-2(x)(3)+3^2}_{(*)}-3^2+6\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\y=(x-3)^2-9+6\\\\y=(x-3)^2-3\to h=3,\ k=-3[/tex]