Johanna spun a spinner 66 times and recorded the frequency of each result in the table. What is the theoretical probability of spinning an odd number? Write your answer using a / to represent the fraction bar.

Answer:

B. -2

Step-by-step explanation:

To find a slope of a line, you have to identify x and y variables:

We use the formula;

m = ∆y/∆x

= (y2 - y1)/(x2 - x1)

Substitute the values into the formula above.

m = (7 - 3)/(2-4)

= 4/-2

= -2

Therefore, the slope of the line is passing through points (4,3) and (2,

7) is -2.

Percent of the original volume that is removed is; 18%

Formula for volume of a box is;

V = lbh

where;

l is length

b is breadth

h is height

Thus;

V_original = 15 * 10 * 8

V_original = 1200 cm³

Volume for each cube removed = 3 * 3 * 3 = 27 cm³

Since there are 8 corners on the box, then 8 cubes are removed.

So the total volume removed is; 8 * 27 = 216

Thus;

Percent of the original volume that is removed is;

216/1200 * 100% = 18%

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The system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X is shown in the figure.

Given that the system of equations are shown in given figure.

The first system of equations are

[tex]\begin{aligned}4x+2y-z&=150\\x+y-z&=-100\\-3x-y+z&=600\\\end[/tex]

By writing in matrix AX=b, we get

Coefficient matrix [tex]A=\left[\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right][/tex] and [tex]B=\left[\begin{array}{l}150&-100&600\end{array}\right][/tex]

Firstly, we will find the A⁻¹ by finding the determinant and adjoint of A and divide the adjoint with determinant, we get

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right|\\ &=4(1-1)-2(1-3)-1(-1+3)\\&=4(0)-2(-2)-1(2)\\ &=2\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}0&2&2\\-1&1&-2\\-2&3&2\end{array}\right]^T\\&=\left[\begin{array}{lll}0&-1&-2\\2&1&3\\2&-2&2\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0&-0.5&-0.5\\1&0.5&1.5\\1&-1&1\end{array}\right]\end[/tex]

For a solution Consider [A B] and apply row operations, we get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{lll1}4&2&-1&150\\1&1&-1&-100\\-3&-1&1&600\end{array}\right]\\ R_{2}&\rightarrow 4R_{2}-R_{1},R_{3}\rightarrow 4R_{3}+3R_{1}\\ &\sim \left[\begin{array}{lll1}4&2&-1&150\\0&2&-3&-550\\0&2&1&2850\end{array}\right]\\ R_{3}&\rightarrow R_{3}-R_{2}\\ &\sim \left[\begin{array}{llll}4&2&-1&150\\0&2&-3&-550\\0&0&4&3400\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}-250\\1000\\850\end{array}\right][/tex]

The second system of equations are

[tex]\begin{aligned}x+y-z&=220\\5x-5y-z&=-640\\-x+y+z&=200\\\end[/tex]

Similarly, we will find for second system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}1&1&-1\\5&-5&-1\\-1&1&1\end{array}\right|\\ &=1(-5+1)-1(5-1)-1(5-5)\\&=1(-4)-1(4)-1(0)\\ &=-8\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-4&-4&0\\-2&0&-2\\-6&-4&-10\end{array}\right]^T\\&=\left[\begin{array}{lll}-4&-2&-6\\-4&0&-4\\0&-2&-10\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.5&0.25&0.75\\0.5&0&0.5\\0&0.25&1.25\end{array}\right]\end[/tex]

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}1&1&-1&220\\5&-5&-1&-640\\-1&1&1&200\end{array}\right]\\ R_{2}&\rightarrow R_{2}-5R_{1},R_{3}\rightarrow R_{3}+R_{1}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&2&0&420\end{array}\right]\\ R_{3}&\rightarrow 5R_{3}+R_{2}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&0&4&360\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}100\\210\\90\end{array}\right][/tex]

The third system of equations are

[tex]\begin{aligned}2x+2y-z&=290\\x+y-3z&=500\\x-y+2z&=600\\\end[/tex]

Similarly, we will find for third system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}2&2&-1\\1&1&-3\\1&-1&2\end{array}\right|\\ &=2(2-3)-2(2+3)-1(-1-1)\\&=2(-1)-2(5)-1(-2)\\ &=-10\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-1&-5&-2\\-3&5&4\\-5&5&0\end{array}\right]^T\\&=\left[\begin{array}{lll}-1&-3&-5\\-5&5&5\\-2&4&0\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.1&0.3&0.5\\0.5&-0.5&-0.5\\0.2&-0.4&0\end{array}\right]\end[/tex]

get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}2&2&-1&290\\1&1&-3&500\\1&-1&2&600\end{array}\right]\\ R_{2}&\rightarrow 2R_{2}-R_{1},R_{3}\rightarrow 2R_{3}-R_{1}\\ &\sim \left[\begin{array}{llll}2&2&-1&290\\&0&-5&710\\0&-4&5&910\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}479\\-405\\-142\end{array}\right][/tex]

Hence, each system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X.

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From the proof of modular congruence below, it has been shown that;

41 ≡ 21 (mod 3).

We want to use the definition of modular congruence to prove that;

41 is congruent to 21 (mod 3) i.e if a ≡ b (mod m) then b ≡ a (mod m).

We are trying to prove that modular congruence mod 3 is a symmetric relation on the integers.

First, if we recall the definition of modular congruence:

For integers a, b and positive integer m,  

a ≡ b (mod m) if and only if m|a–b

Suppose 41 ≡ 21 (mod 3).

Then, by definition, 3|41–21, so there is an integer k such that 41 – 21 = 3k.

Thus;

–(41 – 21) = –3k

So

21 – 41 = 3(–k)

This shows that 3|21 – 41.

Thus;

21 ≡ 41 (mod 3) and the proof is complete

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The value of the indefinite integral is

Given:- The integration of   is given whose lower limit is  and upper limit is .

To Find:- We have to find the value of the integration of   is given whose lower limit is  and upper limit is .

By using the concept of indefinite integral it will be solved.

According to the problem,

Therefore, the value of the indefinite integral is .

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Check the picture below.

part A

since the base of the triangular base is 16, and the altitude "h" splits the base in two equal halves, half that is just 8, so we're looking at a right triangle with a hypotenuse of 17 and a side of 8, thus

[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2 - a^2}=b \qquad \begin{cases} c=\stackrel{hypotenuse}{17}\\ a=\stackrel{adjacent}{8}\\ b=\stackrel{opposite}{h}\\ \end{cases} \\\\\\ \sqrt{17^2-8^2}=h\implies \sqrt{225}=h\implies \boxed{15=h}[/tex]

part B

well, the prism is simply two triangles and 3 rectangles, le's simply add their areas.

[tex]\stackrel{two~triangles}{2\left[ \cfrac{1}{2}(\stackrel{base}{16})(\stackrel{height}{15}) \right]}~~ + ~~\stackrel{two~rectangles}{2(20)(17)}~~ + ~~\stackrel{one~rectangle}{(20)(16)} \\\\\\ 240~~ + ~~680~~ + ~~320\implies \text{\LARGE 1240}[/tex]

The Volume of cylinder represented by the option B.

According to the statement

we have given that the volume formula of cylinder is V= (pi)r^2h, and we have to find that the which expression verify the volume formula for the cylinder given in diagram.

So,

We know that height of the cylinder is given by h = 2x + 7 and

radius r = x - 3.

We know that the formula of volume of cylinder is:

Volume of a cylinder = (pi)r^2h

and Substituting the given values in the above formula

And the volume becomes

Volume = (pi)r^2h

Volume = (pi)( x-3 )^2 (2x+7)

Volume = (pi) ( x^2 + 9 - 6x ) (2x+7)

Volume = (pi) ( 2x^3 + 7x^2 +18x +63 - 42x)

So, The Volume of cylinder represented by the option B.

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Question:

The volume of a cylinder is given by the formula V= pi^2h, where r is the radius of the cylinder and h is the height: which expression represents the volume of this cylinder?

a) Volume = (pi) ( 3x^3 + 7x^2 +14x +63 - 42x)

b) Volume = (pi) ( 2x^3 + 7x^2 +18x +63 - 42x)

c) Volume = (pi) ( 7x^3 + 7x^2 +11x +63 - 42x)

d) Volume = (pi) ( 11x^3 + 7x^2 +13x +63 - 42x)

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Answer:

step 2

Step-by-step explanation:

Erin's mistake is in step 2 it should be 8*2 first and 16* 5 which equals to 86

The difference between the two given fractions is  [tex]\frac{x^{2} +4x-1}{x^{2} +2x}[/tex].

Fraction if a part of whole. It is represented in the form of numerator and denominator.

Given that,

[tex]\frac{x+5}{x+2} -\frac{x+1}{x^{2} +2x}[/tex]

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

Take the LCM of the denominators is [tex]x(x+2)[/tex].

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

=  [tex]\frac{x^{2}+5x -x-1}{x^{2} +2x}[/tex]

=   [tex]\frac{x^{2}+4x-1}{x^{2} +2x}[/tex]

Thus, the difference between the two given fractions is  [tex]\frac{x^{2}+4x-1}{x^{2} +2x}[/tex] .

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The given question is not in correct form.

There will be no change to the grass because the exponential of the function remains the same.

According to the questions,

slope = -4 and y-intercept = -5

Equation of straight line y = mx + c

y = -4x - 5

In order to grass line, there will be no change to the grass because the exponential of the function remains the same.

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Using the Factor Theorem to find the function, the correct statement is:

The function is positive on (0,∞).

The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:

[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]

In which a is the leading coefficient.

The described roots means that:

[tex]x_1 = 0, x_2 = x_3 = x_4 = x_5 = 2[/tex]

Hence the function is:

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

(x - 2)^4 is always positive, hence the sign depends on the sign of x, which means that the correct statement is:

The function is positive on (0,∞).

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1) JKLM is a parallelogram, [tex]\overline{KL} \cong \overline{LM}[/tex] (given)

2) JKLM is a rhombus (a parallelogram with a pair of consecutive congruent sides is a rhombus)

Answer:

3x + 2 >= 0

Step-by-step explanation:

Since this is a 4th root, not a cubic root, the radical can only contain 0 or positive numbers. Therefore there is 3x + 2 >= 0.

Answer:

x < - 9

Step-by-step explanation:

- 4x > 36

divide both sides by - 4 , reversing the symbol as a result of dividing by a negative quantity.

x < - 9

[tex]\boldsymbol{\sf{-4x > 36 }}[/tex]

Divide both sides by −4. Since −4 is <0, the inequality direction is changed.

[tex]\boldsymbol{\sf{x < \dfrac{36}{-4} }}[/tex]

Divide 36 by −4 to get −9.

[tex]\boxed{\boldsymbol{\sf{x < -9}}}[/tex]

The vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

Given an equation showing profits of A Christmas vendor as

P=-0.1[tex]g^{2}[/tex]+30g-1200.

We have to find the number of gingerbread houses that the vendor needs to sell in order to earn profit of $665.60 and $1500.

To find the number of gingerbread houses we have to put P=665.60 in the equation given which shows the profit earned by vendor.

665.60=-0.1[tex]g^{2}[/tex]+30g-1200

0.1[tex]g^{2}[/tex]-30g+1200+665.60=0

0.1[tex]g^{2}[/tex]-30g+1865.60=0

Divide the above equation by 0.1.

[tex]g^{2}[/tex]-300g+18656=0

Solving for g we get,

g=[300±[tex]\sqrt{(300)^{2}-4*1*18656 }[/tex]]/2*1

g=[300±[tex]\sqrt{90000-74624}]/2[/tex]

g=[300±[tex]\sqrt{15376}[/tex]]/2

g=(300±124)/2

g=(300+124)/2       , g=(300-124)/2

g=424/2,  g=176/2

g=212,88

Because 212 is much greater than 88 so vendor prefers to choose selling of 88 gingerbread houses.

Put the value of P=1500 in equation P=-0.1[tex]g^{2}[/tex]+30g-1200.

-0.1[tex]g^{2}[/tex]+30g-1200=1500

0.1[tex]g^{2}[/tex]-30g+1500+1200=0

0.1[tex]g^{2}[/tex]-30g+2700=0

Dividing equation by 0.1.

[tex]g^{2}[/tex]-300g+27000=0

Solving the equation for finding value of g.

g=[300±[tex]\sqrt{300^{2} -4*1*27000}[/tex]]/2*1

=[300±[tex]\sqrt{90000-108000}] /2[/tex]

=[300±[tex]\sqrt{-18000}[/tex]]/2

Because [tex]\sqrt{-18000}[/tex] comes out with an imaginary number so it cannot be solved for the number of gingerbread houses.

Hence the vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

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Formula: U*V=R*T

3*1=4*x

3=4x

x=3/4

Hope it helps!

Answer:

[tex]\sf PR =\dfrac{3}{4}[/tex]

Step-by-step explanation:

It two chords or secants intersect inside the circle, then the product of the length of the segments of one chord is equal to the product of the lengths of the segments of the other chords.

 TP * PR = UP * PV

    4 * PR = 3 * 1

          [tex]\sf PR = \dfrac{3}{4}[/tex]

Answer:

(11a^5 - 3b^4) (3b^4 + 11a^5) = 121a^10 - 9b^8

Step-by-step explanation:

I got the 11 part by doing √121 = 11

I got the a^5 by knowing that a needs to have the same exponent both times and 5+5=10. Thats how I got the a^5 part.

Answer: (11a^5 - 3b^4) (3b^4 + 11a^5) = 121a^10 - 9b^8

Answer:

FIRST

To find the volume of rectangular prism is

Volume = Length x Width x Height

Volume = 20cm x 10cm x 13cm

= 40 x 13

= 520cm cubic or cube

520 is the volume of the rectangular prism

the you divide volume of rectangular prism and volume of solid ball.

Correct me if i am wrong guys.

Answer:

It will take approximately take 17 balls to overflow the container

Step-by-step explanation:

Volume of the rectangle = 20x13x10 = 2600[tex]cm^{3}[/tex]

Volume of the water = 20x10x11 = 2200[tex]cm^{2}[/tex]

Amount of empty space = 2600-2200 = 400[tex]cm^{3[/tex]

Solid ball volume = 23[tex]cm^{3}[/tex] each

To find how many balls can overflow the container = 400/23 = 17.39

The expansion of  (x – 3)^5 is     [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

The Binomial Theorem is the method of expanding an expression that has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc. Binomial Expression: A binomial expression is an algebraic expression that contains two dissimilar terms.

Pascal's Triangle is a never-ending equilateral triangle in which the arrays of numbers arranged in a triangular manner. The triangle starts at 1 and continues placing the number below it in a triangular pattern

Using Binomial theorem,

=[tex]\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}\left(-3\right)^i[/tex]

=[tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

Using Pascal Triangle  for (x-1)

                                               1

                                       1    1              

           1    2    1            

        1    3    3    1        

     1    4    6    4    1      

  1    5    10    10    5    1  

1    6    15    20    15    6    1

Accordingly replacing 1 with 3 we get

= [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

Thus the expansion of  (x – 3)^5 is      [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

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Considering the volume of a rectangular prism, since the new volume is greater than the filled volume, the new dimensions would work, and 90.29% of the infuser would be filled.

The volume of a rectangular prism of length l, width w and height h is given by:

V = lwh.

The original dimensions are:

h = 7.9 cm, l = 3 cm, w = 1.5.

Hence the volume in cm³ was:

V = 7.9 x 3 x 1.5 = 35.55 cm³.

The infuser was filled to 80% capacity, hence:

Vf = 0.8 x 35.55 = 28.44 cm³.

For the new dimensions, we have that h = 7 cm, hence the volume is:

Vn = 7 x 3 x 1.5 = 31.5 cm³.

28.44/31.5 x 100% = 90.29%.

Since the new volume is greater than the filled volume, the new dimensions would work, and 90.29% of the infuser would be filled.

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Considering the least common factor of 15 and 18, it is found that they will depart from the central station at the same time at 11 AM.

To find the time that passes between the events happening at the same time, we need to find the least common multiple of the periods.

In this problem, the periods are of 15 and 18, hence their lcm is found as follows:

15 - 18|2

15 - 9|3

5 - 3|3

5 - 1|5

1 - 1

Hence:

lcm(15,18) = 2 x 3 x 3 x 5 = 90 minutes.

They will depart from the central station at the same time in 90 minutes from 9:30 AM, hence at 11 AM.

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Answer:

C

Step-by-step explanation:

The temperature rises by 30 degrees, so lets say that it goes from 0 to 30.

The temperature is now 30. But then, it falls by 30 degrees. It is now 0 again. The quantities even out, or make 0.

Answer:

Step-by-step explanation:

A parallel line will have the same slope as the line on the graph. A perpendicular line will have a slope that is the opposite reciprocal of the slope of the graphed line.

The line on the graph rises 1 grid square for each run of 2 grid squares to the right. Its slope is ...

  m = rise/run = 1/2

The opposite reciprocal of this slope is ...

  -1/(1/2) = -2

A perpendicular line will have a slope of -2.

The slope-intercept form of the equation for a line is ...

  y = mx +b

where the slope is m and b is the y-intercept. For the purpose here, we don't care about the y-intercepts of any of the lines. We only care about the slope: the coefficient of x.

This means the equations we're looking for are of the form ...

  parallel line: y = 1/2x + b

  perpendicular line: y = 2x +b . . . . . for some constant b

Of the equations offered, the only one with an x-coefficient of 1/2 is ...

  y = 1/2x +3

Of the equations offered, the only one with an x-coefficient of -2 is ...

  y = -2x +1

Answer:

Step-by-step explanation:

you are welcome

Answer:

(32÷(10-8)÷2)-3

Step-by-step explanation:

(32÷(10-8)÷2)-3

(32÷2÷2)-3

(16÷2)-3

8-3 = 5

✅

Based on the Venn diagram, the number of employers who ask employees to be skilled in both communication (C) and handling money (M) is equal to 47 employers.

A Venn diagram is a circular graphical tool that is used to graphically show, logically compare and contrast two (2) or more finite data set or samples in a given population.

From the Venn diagram, we can deduce that the number of employers who ask employees to be skilled in both communication (C) and handling money (M) is given by:

C∩M = 22 + 25

C∩M = 47 employers.

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Answer:

113.04 ft^2

Step-by-step explanation:

area of the circle= 3.14*(6)^2

The probability that the child has measles is gotten as; 0.35

F is the event of a child being sick with flu.

M is the event of a child being sick with measles.

A is the event that the doctor finds a rash.

B1 is the event that the child has measles

S is the sick children.

P(R|M) = 0.93.

P(R|F) = 0.09

P(S|F) = 0.95

P(S|M) = 0.05

Thus, the probability that the child has measles is;

P(M|R) = [(0.05 * 0.93)/[(0.05 * 0.93) + (0.95 * 0.09)]

P(M|R) = 0.35

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Answer:

a) x = 1.5 and x = -0.3

b) x = -8 and x = 5

Step-by-step explanation:

a)

The given equation follows the general structure: ax² + bx + c = 0.

Therefore, if a = 5, b = -6, and c = -2, you can substitute the values into the quadratic formula and solve for "x".

b)

Another way of solving polynomials is through factorization. After rearranging the equation to fit the general structure of a quadratic (as seen above), you can factor by asking yourself the question, which 2 numbers multiply to "c" (-40) and add to "b" (3)? The answers will make up your factors.

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\huge\textbf{Equation \#1. }[/tex]

[tex]\mathsf{5x^2 - 6x - 2 = 0}[/tex]

[tex]\huge\textbf{Use the quadratic formula to solve:}[/tex]

[tex]\mathsf{x = \dfrac{-(-6)\pm \sqrt{(-6)^2 - 4(5)(-2)}}{2(5)}}[/tex]

[tex]\huge\textbf{Simplify it: }[/tex]

[tex]\mathsf{x = \dfrac{6 \pm \sqrt{76}}{10}}[/tex]

[tex]\huge\textbf{Simplify that as well:}[/tex]

[tex]\mathsf{x = \dfrac{3}{5} + \dfrac{1}{5}\sqrt{19}\ or\ x = \dfrac{3}{5} + (-\dfrac{1}{5})\sqrt{19}}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{x \approx 1.5 \ or\ x\approx -0.3{}\ }}\huge\checkmark[/tex]

[tex]\huge\textbf{Equation \#2.}[/tex]

[tex]\mathsf{x^2 + 3x = 40}[/tex]

[tex]\huge\textbf{Subtract 40 to both sides:}[/tex]

[tex]\mathsf{x^2 + 3x - 40 = 40 - 40}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\mathsf{x^2+ 3x - 40 = 0}[/tex]

[tex]\huge\textbf{Factor the left side of the equation:}[/tex]

[tex]\mathsf{(x - 5)\times (x + 8) = 0}[/tex]

[tex]\mathsf{(x - 5)(x + 8) = 0}[/tex]

[tex]\huge\textbf{Set the factors to equal to 0:}[/tex]

[tex]\mathsf{x - 5 = 0 \ or\ even\ x + 8 = 0}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\mathsf{x = 5\ or\ x = -8}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{x = 5\ or \ x = -8}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

A 3 1 B 2-4 C4-3 then work x and y graph