Find the area of the shaded region if the dimensions of the unshaded region are 18ft x 22ft. Use 3.14 for π as necessary.
A. 1,419.84 ft²
B. 1,111.84 ft²
C. 709.92 ft²
D. 957.84 ft²

Answer:

The opción b

Step-by-step explanation:

the profit is defined by

[tex]Profit=Gain-cost[/tex]

In this case [tex](F-g)(100)=[/tex] is the profit to sell 100 television and is calculated

[tex]F(x)-g(x)=4x+5000-(20x-500)=4x-20x+5000+500=-16x+5500=-16*(100)+5500=3900[/tex]

[tex]3900=3.9k[/tex]

Answer:

The answer You're looking for is B)$3.9K is the profit from 100 TV's

Step-by-step explanation:

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:

y = 1/2 x + 8

Step-by-step explanation:

We are given the y intercept and the slope, so we can write the equation using the slope intercept form

y = mx+b  where m is the slope and b is the y intercept

y = 1/2 x + 8

Answer:

   Slope of Function B = 2 x Slope of Function A

Step-by-step explanation:

step 1

Find the slope of the function A

we have

[tex]f(x)=x-9[/tex]

This is the equation of the line into point slope form

[tex]y=mx+b[/tex]

where m is the slope

b is the y-intercept

therefore

The slope of the function A is

[tex]m=1[/tex]

step 2

Find the slope of the function B

we have the points

(-1,-3) and (2,3)

The slope m is equal to

[tex]m=(3+3)/(2+1)=6/3=2[/tex]

step 3

Compare the slopes

[tex]SlopeA=1\\ SlopeB=2[/tex]

therefore

The slope of the function B is two times the slope of the function A

Answer:

slope of function a = -2

slope of function b = (1 + 5)/(2 + 1) = 6/3 = 2

slope of function b = - slope of function a.

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]1.\\\text{The length of semicircle:}\\\\l=\dfrac{1}{2}d\pi\\\\d-diameter\\\\d=2.2\\\\\text{substitute:}\\\\l=\dfrac{1}{2}(2.2)\pi=1.1\pi\approx(1.1)(3.14)=3.454\\\\\text{The perimeter of the figure:}\\\\P=2(3.8)+2.2+3.454=13.254[/tex]

[tex]2.\\P=2(3.32)+2x=6.64+2x[/tex]

[tex]6.\\\text{Look at the picture.}\\\\P=4(2)+2(1)=8+2=10[/tex]

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

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!

P(1, 4), Q(6, 4), R(6, 1), and S(1, 1)

New coordinates of the rectangle after a reflection over the x-axis is:

P'(1, -4), Q'(6, -4), R'(6, -1), and S'(1, -1)

New coordinates of the rectangle after a reflection over the y-axis is:

P"(-1, 4), Q"(-6, 4), R"(-6, 1), and S"(-1, 1)

.

Answer:

Over the Y axis.

P(-1, 4), Q(-6, 4), R(-6, 1), and S(-1, 1)

Over the X axis:

P(1, -4), Q(6, -4), R(6, -1), and S(1, -1)

Step-by-step explanation:

In order to finde the new coordinates after a reflection on the Y or X axis, you just have to change the sign of the opposite variable, for example if you want to reflect the points on the Y axis you change the signs of all the X´s on the points, and the same with the reflections of the X axis, you have to change the signs of all the Y´s on the points.

Answer:

The answer is FEG and EGF

The pair of angles, FEG and FGE, are nonadjacent and complementary, which means their sum is 90° but they do not have a common side.

Two angles are said to be complementary if their sum is 90° and two angles are supplementary if their sum is 180°.

We know two angles are complementary when their sum is 90°, and we also know that two angles are adjacent if they share a common side.

Here we have a pentagon with five sides.

The pair of angles are nonadjacent and complementary meaning their sum is 90° but they do not share a common side they are,

∠FEG and ∠FGE, because ∠FEG + ∠FGE = 55° + 35° = 90°.

learn more about complementary angles here :

https://brainly.com/question/15592900

#SPJ5

The approximate circumference of a circle with a diameter of 9 is 28.3 units, rounded to the nearest tenth, calculated using the formula C = πd.

The circumference of a circle can be calculated using the formula C = πd, where C is the circumference and d is the diameter of the circle. Given the diameter of 9, we can substitute this value into the formula to calculate the circumference.

Therefore, the circumference C is:

C = π × 9

Using the approximation for π as 3.14, we get:

C ≈ 3.14 × 9 = 28.26

So, the approximate circumference of the circle is 28.3 units, rounded to the nearest tenth.

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.

learn more about test here: https://brainly.com/question/20416466

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:

Option B

Step-by-step explanation:

the mean is the sum of each value of P(x) multiply for x

Mean= 23x0.16 + 25x0.09 + (26x0.18) + (31x0.12)+ (34x0.24) + (38x0.21)

Mean= 30.47

Answer:

B. 30.47

Step-by-step explanation:

The mean discrete random variable tells us the weighted average of the possible values given of a random variable. It shows the expected average outcome of observations. It's like getting the weighted mean. The expected value of X can be computed using the formula:

[tex]\mu_{x}=x_1p_1+x_2p_2+x_3p_3...+x_kp_k\\\\=\Sigma x_ip_i[/tex]

Using the data given in your problem, just pair up the x and P(x) accordingly and get the sum.

[tex]\mu_{x}=x_1p_1+x_2p_2+x_3p_3+x_4p_4+x_5p_5+x_6p_6\\\\\mu_{x}=(23\times 0.16) + (25\times 0.09) + (26\times 0.18) + (31\times 0.12) + (34\times 0.24) + (38\times 0.21)\\\\\mu_{x} = 30.47[/tex]

5

Step-by-step explanation:

You can use Tan(45)=S/5 to solve for S. Tan basically means Opposite/Adjacent. So the opposite side is S and the adjacent side is S. You plug tan(45) into your calc and then multiply it by 5.

Proportion Below

Tan(45)/1 = S/5

You cross multiply to get Tan(45)*5=S

S would be 5

To check you can use the Pythagorean theorem a^2+b^2=c^2

5^2+5^2=

[tex] {5}^{2} + {5}^{2} = \sqrt{50} [/tex]

25+25=50

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:

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.

https://brainly.com/question/34235911

Answer:

[tex]\frac{5\pi }{18}[/tex]

Step-by-step explanation:

To convert from degree to radian measure

radian measure = degree measure × [tex]\frac{\pi }{180}[/tex]

For 50°, then

radian = 50 × [tex]\frac{\pi }{180}[/tex]

Cancel the 50 and 180 by 10, leaving

radian measure = [tex]\frac{5\pi }{18}[/tex]

Answer:

3/10

Step-by-step explanation:

These two events are independent, so the overall probability is the product of the individual probabilities.

12 blue sections out of 20

p(blue) = 12/20 = 3/5

There is an equal probability of the coin landing on heads or tails.

p(tails) = 1/2

p(blue & tails) = 3/5 * 1/2 = 3/10

See the attached picture:

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

Answer with step-by-step explanation:

1) Inverse of [tex]f(x) = 6x -24[/tex]:

Make the function equal to y to get [tex]y=6x-24[/tex].

Now making [tex]x[/tex] the subject of the function:

[tex]6x=y+24\\\\x=\frac{y+24}{6} \\\\x=\frac{y}{6} +4[/tex]

Change back the variable [tex]y[/tex] to [tex]x[/tex] and this is the inverse.

[tex]f'(x)=\frac{x}{6} +4[/tex]

2. Inverse of [tex]g(x) = 3x^2 - 5[/tex]:

Making the function equal to y to get: [tex]y=3x^2 - 5[/tex]

Now making [tex]x[/tex] the subject of the function:

[tex]3x^2=y+5\\\\x^2=\frac{y+5}{3}[/tex]

Taking square root on both sides to get:

[tex]x=\sqrt{\frac{y+5}{3} }[/tex]

Change back the variable [tex]y[/tex] to [tex]x[/tex] and this is the inverse.

[tex]g'(x)=\sqrt{\frac{x+5}{3} }[/tex]

Answer:

1) [tex]f^{-1}(x)=\frac{x}{6}+4[/tex]

2)  [tex]g^{-1}(x)=\sqrt{\frac{x+5}{3}}[/tex]

Step-by-step explanation:

1) To find the inverse of the function [tex]f(x) = 6x -24[/tex] you need to follow these steps:

- Since [tex]f(x)=y[/tex], you can rewrite the function:

 [tex]y = 6x -24[/tex]

- Solve for "x":

[tex]y+24=6x\\\\x=\frac{y+24}{6}\\\\x=\frac{y}{6}+4[/tex]

- Exchange the variables:

[tex]y=\frac{x}{6}+4[/tex]

Then, the inverse is:

[tex]f^{-1}(x)=\frac{x}{6}+4[/tex]

2) To find the inverse of the function [tex]g(x) = 3x^2 - 5[/tex] you need to follow these steps:

- Since [tex]g(x)=y[/tex], you can rewrite the function:

 [tex]y= 3x^2 - 5[/tex]

- Solve for "x":

[tex]y= 3x^2 - 5\\\\\frac{y+5}{3}=x^2\\\\x=\sqrt{\frac{y+5}{3}}[/tex]

- Exchange the variables:

[tex]y=\sqrt{\frac{x+5}{3}}[/tex]

 Then, the inverse is:

[tex]g^{-1}(x)=\sqrt{\frac{x+5}{3}}[/tex]

Answer: A is correct, (x-5, y-3) Hope this helps, mark brainliest please :)

Step-by-step explanation:

The green triangle is the original, and the blue is the duplicated translation, you know this by the ' mark on the blue S.

You can count the squares from one spot to the next to find how many it moved. It clearly is moved to the left and down though, which means it was subtracted from in both directions.

Answer:A

Step-by-step explanation:

Answer:

9x^8

Step-by-step explanation:

A perfect square is a number that has a whole number square rooth. Therefore we need to find two numbers 'a' and 'b' that satisfy the following equation:

3x^4 = sqrt(ax^b)

(3x^4)^2 = ax^b

9x^8 = ax^b

a=9 and b=8.

Therefore, the term 9x^8 is a perfect square of the root 3x^4.

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:

6.95 pounds. Her dog weighs 6.95 more pounds than the total weight of both her cats.

Step-by-step explanation:

To find the total weight of her cats, add their weights together:

9.8+8.25

=18.05 pounds

Then to find the difference between the weight of her dog and the weight of her cats' total weight, you subtract the weight of the cats from the weight of the dog:

25-18.05

=6.95 pounds

[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]

Answer:

It's B. 40,653

Step-by-step explanation:

Answer:b

Step-by-step explanation:

Answer:

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

Then, the work consists in finding pairs (a,b) for which:

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

[tex]\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}[/tex]

I will use b =0.5, b = 1, b = 1.5, b = 2

[tex]b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236[/tex]

Then, four distinct complex numbers that have an absolute value of 3 are:

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:

(-0.23,5.77)

Step-by-step explanation:

The given system of equations is

[tex]y = - 34x - 2...(1)[/tex]

[tex]y = x + 6...(2)[/tex]

We equate both equations to get:

[tex]x + 6 = - 34x - 2[/tex]

Group similar terms to get:

[tex]x + 34x = - 2 - 6[/tex]

[tex]35x = - 8[/tex]

[tex]x = - \frac{8}{35} [/tex]

[tex]x \approx - 0.23[/tex]

Put this value into the second equation to get y

[tex]y = - 0.23 + 6 \approx 5.77[/tex]

(-0.23,5.77)