Which set of ordered pairs does not represent a function?
O {(0,6), (5,-4), (3, -4), (8,-6)}
O {(6,-1), (-6, -1),(-2,-8),(-4,9)}
O {(-3, -3), (4,1),(-3,3), (2, 2)}
O {(1,-4), (5, -7), (-3,-6), (-2,-4)}

Yes, because both figures are rectangles and all rectangles are similar. Option B is correct option.

According to the statement

We have given that the two rectangles EFGH and ABCD and we have show that the these rectangles Are dilation at origin or not.

So, According to the given diagram

Yes, the rectangle EFGH is a result of dilation of rectangle ABCD with a center of dilation of rectangle at the origin.

Also The scale factor of the dilation is greater than one as the image is bigger than the pre-image i.e. there is a stretch.

The scale factor could be calculated by the ratio of the sides of the image to the pre-image rectangle.

According to the diagram In Rectangle ABCD:

Its vertices have coordinates A(-3,3), B(3,3), C(3,0) and D(-3,0).

Now consider rectangle EFGH:  

Its vertices have coordinates E(-4,4), F(4,4), G(4,0) and H(-4,0).

Hence the scale factor is becomes from these values is:

EF/AB = EH/AD = FG/BC = HG/DC = 4/3.

Hence the scale factor becomes 4/3.

Also

∠A=∠B=∠C=∠D=∠E=∠F=∠G=∠H=90°

( When a shape is dilated the two shapes are similar.And similar shapes have equal interior angles , corresponding sides are proportional ).

So, Yes, because both figures are rectangles and all rectangles are similar. Option B is correct option.

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Monthly interest payment is $100.

According to the statement

balance in account = $100

APR = 12%

we use the formula Balance formula to solve this problem.

So, BALANCE * [APR / 12 month] to find monthly interest payment

Here balance is the starting amount in the bank account

And APR is the type of interest rate applied on the amount

And 12 month is the time period for which APR is applicable.

So, substitute the values in it then

Monthly interest payment = 100 * [12/12]

After solving the equation become

Monthly interest payment = 100*1

Monthly interest payment = 100$

So, Monthly interest payment is $100.

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Answer:

-1/2

Step-by-step explanation:

cos(x+pi) = -cos(x), so the answer is -1/2.

The exponential function that models this situation is:

[tex]D(t) = 10000(1.045)^t[/tex]

Using the function, in 2050, she will have $37,453.

An increasing exponential function is modeled by:

[tex]A(t) = A(0)(1 + r)^t[/tex]

In which:

For this problem, the parameters are:

A(0) = 10000, r = 0.045.

Hence the equation is:

[tex]D(t) = 10000(1.045)^t[/tex]

2050 is 30 years after 2020, hence the amount is:

[tex]D(30) = 10000(1.045)^{30} = 37453[/tex]

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The volume of the prism would be enlarged by a factor of 8

The scale factor is given as:

k = 2

Let the initial volume be v and the final volume be V.

The relationship between both volumes is

V = k^3 * v

This gives

V = 2^3 * v

Evaluate

V = 8v

Hence, the volume of the prism would be enlarged by a factor of 8

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Answer:

Zero real number solutions

Step-by-step explanation:

Doesn't intercept the x-axis -> No real solutions

The total possible number of outcomes for the season record is [tex]2^{33}[/tex].

An event that may be given a mathematical probability is known as an outcome in mathematics (an outcome of an experiment). The amount of potential outcomes for a particular experiment determines the probability assigned to each result.The result of an experiment is a number of outcomes.

According to the question,

The basketball team at a college will play 33 games next winter. There are only two possible outcomes for each game: a win or a loss.

So, the total number of possible outcomes for the season record=[tex]2^{33}[/tex].

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1. 1/2

2.2/3

3. 1

4. 1/3

5. 2/7

6. 1/4

7. 4/3

8. 3/2

To find 2x of the vale of x, we have to multiply the value of 'x' by 2

1. x = 1/4

[tex]2x = 2 * \frac{1}{4}[/tex] ⇒ [tex]\frac{2}{4}[/tex] ⇒[tex]\frac{1}{2}[/tex]

2. x = 1/3

[tex]2x = 2 * \frac{1}{3}[/tex] ⇒[tex]\frac{2}{3}[/tex]

3. x = 1/2

[tex]2x = 2* \frac{1}{2}[/tex] ⇒ [tex]\frac{2}{2}[/tex] ⇒ [tex]1[/tex]

4. x= 1/6

[tex]2x = 2* \frac{1}{6}[/tex] ⇒ [tex]\frac{2}{6}[/tex] ⇒ [tex]\frac{1}{3}[/tex]

5. x = 1/ 7

[tex]2x = 2 * \frac{1}{7}[/tex] ⇒ [tex]\frac{2}{7}[/tex]

6. x = 1/8

[tex]2x = 2 * \frac{1}{8}[/tex] ⇒ [tex]\frac{2}{8}[/tex] ⇒ [tex]\frac{1}{4}[/tex]

7. x = 2/3

[tex]2x = 2 * \frac{2}{3}[/tex] ⇒ [tex]\frac{4}{3}[/tex]

8. x = 3/4

[tex]2x = 2 * \frac{3}{4}[/tex] ⇒ [tex]\frac{6}{4}[/tex] ⇒ [tex]\frac{3}{2}[/tex]

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A paragraph proof D. contains a set of sentences explaining the steps needed to reach a conclusion.

It should be noted that a paragraph proof simply means a way of presenting a mathematical proof.

In this case, it contains a set of sentences explaining the steps needed to reach a conclusion.

In conclusion, the correct option is D.

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Answer:

D: contains a set of sentences explaining the steps needed to reach a conclusion.  

Answer:

  35:9

Step-by-step explanation:

The ratios can be combined by making the common element be represented by the same number.

The common element in the two ratios is the number of girls. In the first ratio, girls are represented by 3 ratio units. In the second ratio, they are represented by 7 ratio units. Multiplying the first ratio by 7 and the second ratio by 3 will make the number of girls the same in both.

  boys : girls = 2 : 3 = 14 : 21

  girls : teachers = 7 : 3 = 21 : 9

Then the extended ratio can be written ...

  boys : girls : teachers = 14 : 21 : 9

The number of students is the total of the numbers of boys and girls, so we have ...

  students : teachers = (14 +21) : 9

  students : teachers = 35 : 9

Based on the number of shares that Ralph Warren purchased, the total cost of the stock was $469.63. The amount he received from sales was $447.13. The capital loss was $22.50.

The cost of the stock was:

= (27 x 16³/₈) + 27.50

= $469.63

The amount received from sales:

= (27 x 17⁵/₈) - 28.75

= $447.13

This capital gain (loss):

= 447.13 - 469.63

= -$22.50

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Answer: 2400

Step-by-step explanation:

3000 x 20/100 = 600

Discount = 600

Sale price of the article: 3000-600 = 2400

[tex] \underline{ \orange{\huge \boxed{ \frak{Answer : }}}}[/tex]

Let ,

[tex] \sf \large \color{purple} y = 2 {x}^{3} - 3 {x}^{2} - 12x + 10 \: --( \: 1 \: )[/tex]

[tex] \: \: \: [/tex]

Now , Diff wrt ' x ' , we get :

[tex] \sf \: \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{3} - 3 {x}^{2} - 12x + 10) \\ \sf \: \sf \: \frac{dy}{dx} = \frac{d}{dx} \: 2(3 {x}^{2} ) - \frac{d}{dx} 3 {x}^{2} - \frac{d}{dx} 12x + \frac{d}{dx} 10 \\ \sf \: \frac{dy}{dx} =2(3 {x}^{2} ) - 3(2x) - 12(1) + 0 \\ \sf \: \frac{dy}{dx} =6 {x}^{2} - 6x - 12 + 0 \\ \: \sf \red{\frac{dy}{dx} = 6 {x}^{2} 6x - 12 -- (2)}[/tex]

[tex] \: \: \: [/tex]

For maxima or minima \frac{dy}{dx} = 0

[tex] \: \: \: [/tex]

[tex] \sf \: 6 {x}^{2} - 6x - 12 = 0[/tex]

[tex] \: \: \: [/tex]

Divided by 6 on both side , we get.

[tex] \: \: \: [/tex]

[tex] \sf \: {x}^{2} - x - 2 = 0 \\ \sf \: {x}^{2} - 2x + x - 2 = 0 \\ \sf \: x(x - 2) + 1(x - 2) = 0 \\ \sf \: (x - 2)(x + 1) = 0 \\ \sf \: x - 2 = 0 \: \: \bold or \: \: x + 1 = 0 \\ \sf \fbox{x = 2 \: } \: \bold or \: \fbox{ x = - 1}[/tex]

[tex] \: \: \: [/tex]

Again Diff wrt ‘ x ’ , we get.

[tex] \sf \: \frac{d}{dx} =(\frac{dy}{dx} ) = 6\frac{d}{dx} - 6\frac{d}{dx}x - \frac{d}{dx}12 \\ \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = 6(2x) - 6(1) - 0 \\ \sf \: \sf \bold{ \frac{ {d}^{2}y }{ {dx}^{2} } =12x - 6}[/tex]

[tex] \: \: \: [/tex]

At x = 2

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12(2) - 6 \\ \: \: \: \sf \: = 24 - 6 \\ \: \: \: \: \sf \red{ = 18 > 0}[/tex]

At x = -1

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12( - 1) - 6 \\ \: \: \: \sf \: = - 12 - 6 \\ \: \: \: \: \sf \red{ = - 18 < 0 }[/tex]

[tex] \: \: \: [/tex]

x = 2 gives minima value of function.

[tex] \: \: \: [/tex]

x = -1 gives maxima value of function.

[tex] \: \: \: [/tex]

Now, put x = 2 in eqⁿ ( 1 )

[tex] \: \: \: [/tex]

[tex] \sf \: y \: minima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( {2})^{3} - 3 ({2})^{2} - 12(2) + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: = 2(8) - 3(4) - 24 + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: = 16 - 12 - 24 + 10 \\\sf \: \: \: \: \: \: \: \: \: \: = - 20 + 10 \\\sf \color{red}{\boxed{ = - 10}}[/tex]

[tex] \: \: \: [/tex]

The Point of minima is ( 2 , -10 ).

[tex] \: \: \: [/tex]

Now , put x = -1 in eqⁿ ( 1 )

[tex]\sf \: y \: maxima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( { - 1})^{3} - 3 ({ - 1})^{2} - 12( - 1) + 10 \\\sf \color{red}{\boxed{ = 17}}[/tex]

[tex] \: \: \: [/tex]

The point of maxima value is ( -1 , 17 ).

[tex] \: \: \: [/tex]

[tex] \: \: [/tex]

Hope Helps! :)

Using the binomial distribution, it is found that the standard deviation of the distribution is of 0.67.

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

The parameters are:

n = 5, p = 0.1.

Hence the standard deviation is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5 \times 0.1 \times 0.9} = 0.67[/tex]

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Answer:

d

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

(x - 5 )² + (y + 1)² = 4 ← is in standard form

with centre (5, - 1 ) and r = [tex]\sqrt{4}[/tex] = 2

Option B. The situation that would be more likely to show a constant rate of change would be B. The distance traveled by a truck driving at a constant speed compared with time.

This is the factor that is used to determine if there is a decrease or an increase in a given factor.

A good way to d this is by the use of time and speed to compare the distance that is being traveled.

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Complete question

Which situation is most likely to show a constant rate of change?

A. The weight of a child compared with its age in months

B. The distance traveled by a truck driving at a constant speed compared with time

C. The cost of a new car compared with the number of tires it has

D. The outside temperature compared with the time of day

The equation of the line of best fit is y = 1.7x  - 58

We start by drawing the line of best fit (see attachment)

From the attached graph, we have the following points

(x, y) = (70, 75) and (61, 60)

The slope (m) is:

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

This gives

m = (60 - 75)/(61 - 70)

Evaluate

m = 1.7

The line of best fit is then calculated as:

y = m(x - x1) + y1

This gives

y = 1.7(x - 70) + 61

This gives

y = 1.7x - 119 + 61

Evaluate

y = 1.7x  - 58

Hence, the equation of the line of best fit is y = 1.7x  - 58

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The exponent Marina should use for b is: D. 7.

When dividing, you subtract the exponents. I.e. [tex]a^m \div a^n = a^{m - n}[/tex].

Using the rule, we would also subtract the exponents of b in the simplified expression. Thus:

[tex]b^4 \div b^{-3}[/tex]

Subtract the exponents

4 -(-3) = 4 + 3 = 7.

Therefore, the Maria should use the exponent b = 7.

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Answer:  D

Step-by-step explanation:

Answer:

D.

[tex] \frac{1}{16?} [/tex]

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

[tex]\huge\textbf{Equation: }[/tex]

[tex]\mathbf{2^{-4}}[/tex]

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

[tex]\mathbf{2^{-4}}[/tex]

[tex]\mathbf{\approx \dfrac{1}{2^4}}[/tex]

[tex]\mathbf{= \dfrac{1}{2\times2\times2\times2}}[/tex]

[tex]\mathbf{= \dfrac{1}{4\times4}}[/tex]

[tex]\mathbf{= \dfrac{1}{16}}[/tex]

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

[tex]\huge\boxed{\frak{Option\ D. \ \dfrac{1}{16}}}\huge\checkmark[/tex]

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

Answer:

First function:

Zeros = 6 and -6. Y-intercept = (0,72). X-intercepts = (6,0) and (-6,0).

Second function:

Zeros = 3 and -3. Y-intercept = (0, -27). X-intercepts: (3, 0) and (-3, 0).

Step-by-step explanation:

The zeros are x-intercept numbers.

The y-intercept can be found when the function is in standard form. Plug in 0 for x, then solve.

Hope this helps!

Answer:

1. x₁ = 6, x₂ = -6

2. x₁ = 3, x₂ = -3

Step-by-step explanation:

Given functions:

[tex]1)\ f(x) = -2(x-6)(x+6)[/tex]

[tex]2)\ f(x)=(x-3)(3x+9)[/tex]

..................................................................................................................................................

Zero Product Property: If m • n = 0, then m = 0 or n = 0.

Standard Form of a Quadratic: ax² + bx + c = 0.

..................................................................................................................................................

1. f(x) = -2(x - 6)(x + 6)

Step 1: Set the function to zero.

[tex]\implies 0 = -2(x-6)(x+6)[/tex]

Step 2: Divide both sides of equation by [tex]-2[/tex].

[tex]\implies \dfrac{0}{-2} = \dfrac{-2(x-6)(x+6)}{-2}[/tex]

[tex]\implies 0=(x-6)(x+6)[/tex]

Step 3: Apply the Zero Product Property.

[tex]x_1 \implies x-6=0[/tex]

[tex]x_2 \implies x+6=0[/tex]

Step 4: Solve for x in both equations.

[tex]x-6+6=0+6 \implies \boxed{x_1 = 6}[/tex]

[tex]x+6-6=0-6 \implies \boxed{x_1 = -6}[/tex]

The zeros (x-intercepts) of this function are: [tex]x_1=6,\ x_2=-6[/tex].

.................................................................................................................................................

2. f(x) = (x - 3)(3x + 9)

Step 1: Set the function to zero.

[tex]\implies 0 = (x - 3)(3x + 9)[/tex]

Step 2: Apply the Zero Product Property.

[tex]x_1 \implies x-3=0[/tex]

[tex]x_2 \implies 3x + 9=0[/tex]

Step 3: Solve for x in both equations.

[tex]x-3+3=0+3 \implies \boxed{x_1 = 3}[/tex]

[tex]3x=-9 \implies \dfrac{3x}{3}=\dfrac{-9}{3} \implies \boxed{x_1 = -3}[/tex]

The zeros (x-intercepts) of this function are: [tex]x_1=3,\ x_2=-3[/tex].

..................................................................................................................................................

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The other factor of the polynomial is a. 3x2-4x+11

Given polynomial is 3x3 + 20x2 - 21x + 88 and the factor is (x+8)

We need to find another factor using long division method

So,

We will divide the polynomial by the factor to find the another factor

Therefore,

[tex]\sqrt[x+8]{3x^3+20x^2-21x+88}[/tex]

Now calculating

First multiplying [tex]3x^2[/tex] with (x+8) so , [tex]3x^2[/tex] will be in the quotient

We get [tex]3x^3+24x^2[/tex]

simplifying the calculation for [tex]3x^3+20x^2[/tex] and [tex]3x^3+24x^2[/tex]

We get the remainder is [tex]-4x^2-21x+88[/tex]

Second we will multiply -4x with (x+8) Where -4x will be in the quotient

We get [tex]-4x^2-32x[/tex] and then we will simplify the equation

We get 11x +88 as a remainder

The quotient we get is [tex]3x^2-4x[/tex]

Third we will multiply +11 with (x+8) Where +11 will be in the quotient

we get 11x+88

Simplifying the equation we get the remainder 0

So the quotient we get is  (3x2 - 4x+11)

Hence the another factor of the polynomial is  (3x2 - 4x+11)

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Answer:

5

Step-by-step explanation:

(2x - 6) + 1 + (x - 4) = 18

3x - 9 = 18

3x = 27

x = 9

RS = x - 4 = 9 - 4 = 5

Answer:

B. 5

Step-by-step explanation:

Pls mark brainiest if helpful!!

PQ + QR + RS = 18

This statement is equivalent to the following:

(2x - 6) + 1 + (x - 4) = 18.

This is because we substitute the line segments with their values.

Now that we have an equation to work with, we will solve it. We start by removing the parenthesis as they are not necessary here.

2x - 6 + 1 + x - 4 = 18

Next, we will make it, so x is alone on the left. We will do this by adding 6 and 4 to both sides

2x - 6 (+6) + 1 + x - 4 (+4) = 18 (+6) (+4)

After doing this we are left with the following equation:

2x + 1 + x = 28

Now we will subtract 1 from both sides to make x alone of the left side of the equation.

2x + 1 (-1) + x = 28 (-1)

After we solve this, we get the following:

2x + x = 27

We can add 2x and x and get 3x because 2 + 1 = 3.

3x = 27

We now simplify this equation by dividing both sides by 3.

[tex]\frac{3x}{3}[/tex] = [tex]\frac{27}{3}[/tex]

We again simplify the equation to the following:

x = 9

BUT THATS NOT THE ANSWER!

Now that we know what x is, we can solve for the value of RS.

We input 9 for x in the equation x - 4.

When we do this, we get 9 - 4 which is 5.

And this is our final answer!!


I hope I could help! Have an amazing day and good luck on your homework!

"You can't predict the future, but you can create it"
            -Juliana Palazzolo, 2022

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Answer: c

Step-by-step explanation:

The equation of the line is

[tex]y+2=\frac{1}{3}(x-1)\\\\y+2=\frac{1}{3}x-\frac{1}{3}\\\\y=\frac{1}{3}x-\frac{7}{3}[/tex]

To determine which point lies on the line, we can substitute in the x coordinate and see if the equation gives the same y-coordinate.

Therefore, the answer is c.

Find y intercept

Point(0,-7)

Find vertex

x coordinate

y coordinate

Vertex at (-2,-27)

Slope

Answer:

10

Step-by-step explanation:

The x-coordinate of the vertex of a quadratic equation in the form

[tex]f(x)=ax^2+bx+c\quad \textsf{is} \quad -\dfrac{b}{2a}[/tex]

Given function:

[tex]f(x)=5x^2+20x-7[/tex]

[tex]\implies a=5, \quad b=20, \quad c=-7[/tex]

x-coordinate of the vertex

[tex]\implies -\dfrac{b}{2a}=-\dfrac{20}{2(5)}=-2[/tex]

To find the y-coordinate of the vertex, substitute the found value of x into the function:

[tex]\begin{aligned}\implies f(-2) & =5(-2)^2+20(-2)-7\\& = 5(4)-40-7\\& = 20-47\\& = -27\end{aligned}[/tex]

Therefore, the coordinates of the vertex are (-2, -27).

The y-intercept is when the curve crosses the y-axis, so when x = 0.

To find the y-coordinate of the y-intercept, substitute x = 0 into the function:

[tex]\begin{aligned}\implies f(0) & =5(0)^2+20(0)-7\\& = 0 + 0-7\\& = -7\end{aligned}[/tex]

Therefore, the coordinates of the y-intercept are (0, -7).

To find the slope of the line passing through the vertex and the y-intercept, simply substitute the found points into the slope formula:

[tex]\implies \sf slope=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{-27-(-7)}{-2-0}=\dfrac{-20}{-2}=10[/tex]

Therefore, the slope of the line passing through the vertex and the y-intercept of the given quadratic function is 10.

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Question 4

1) [tex]\angle B[/tex] and [tex]\angle E[/tex] are right angles, [tex]\overline{AB} \cong \overline{DE}[/tex], [tex]\angle A \cong \angle D[/tex] (given)

2) [tex]\triangle CBA[/tex] and [tex]\triangle FED[/tex] are right triangles (a triangle with a right angle is a right triangle)

3) [tex]\triangle CBA \cong \triangle FED[/tex] (LA)

Question 5

1) [tex]\overline{XY} \perp\overline{WZ}[/tex], [tex]\overline{UV} \perp \overline{WZ}[/tex], [tex]\overline{VW} \cong \overline{YZ}[/tex], and [tex]\angle Z \cong \angle W[/tex] (given)

2) [tex]\angle XYZ[/tex] and [tex]\angle UVW[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle XYZ[/tex] and [tex]\triangle UVW[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\triangle XYZ \cong \triangle UVW[/tex] (LA)

5) [tex]\overline{UW} \cong \overline{XZ}[/tex] (CPCTC)

Question 6

1) [tex]\overline{PQ} \perp \overline{QT}[/tex], [tex]\overline{ST} \perp \overline{QT}[/tex], [tex]\overline{PQ} \perp \overline{ST}[/tex] (given)

2) [tex]\angle PQR[/tex] and [tex]\angle RTS[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle PQR[/tex] and [tex]\triangle STR[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\angle PRQ \cong \angle SRT[/tex] (vertical angles are congruent)

5) [tex]\triangle PQR \cong \triangle STR[/tex] (LA)

6) [tex]\overline{QR} \cong \overline{TR}[/tex] (CPCTC)

The plane is parallel to the base of the cone describes the relationship between the plane and the cone

A circle is a spherical form without boundaries or edges or we can say that A circle is a closed, curved object with two dimensions in geometry.

The produced cross-section is a circle when a plane slices a cone such that the plane is parallel to the base of the cone.

Ellipse is the shape that is produced when a plane cuts a cone at an angle such that it misses the base or the vertex. (As shown in figure)

Hence A circle is formed when a plane slices through a cone and the plane is parallel to the base of the cone

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Answer:

13 + [tex]\sqrt{89}[/tex]

or

22.43398... (depends on how you round it)

Step-by-step explanation:

1. The question stated that the radius of the circle O is 5, so the length of AO and CO is 5.

2. Since line AB and line CB are both tangent to the circle, they have the same length. CB is 8, so AB will also be 8.

--> Both triangle AOB and COB share one side, and the other side (radius) has the same length, so the third side must be the same length

3. Tangent means having a 90-degree angle with the radius. We know that the triangle AOB is a right triangle since the angle OAB is 90 degrees.

We can use the Pythagorean theorem to find side OB. OB^2 = AO^2 + AB^2.

--> OB^2 = 25 + 64

--> OB^2 = 89

--> OB = [tex]\sqrt{89}[/tex]

Now that we know the lengths of all three sides, we can add them up.

--> 5 + 8 + [tex]\sqrt{89}[/tex]

--> 13 + [tex]\sqrt{89}[/tex]

or

--> 22.43398113....

Answer:

4x^3 + 7x^2 - 3x + 1

Step-by-step explanation:

BTW your question, there is a sign misssing

Answer is 5/4The slope is the number next to the “x” in the y intercept formula

y= m (slope) x + b

Answer: m= 5/4

Step-by-step explanation:

La ecuación general de la recta es y=mx+by=mx+b, donde m es la pendiente y b es la intersección en y

y=mx+by=mx+b