If AABC = ADEF, which congruences are true by CPCTC? Check all that
apply.

(a) If [tex]\alpha[/tex] and [tex]\beta[/tex] are roots of [tex]f(x)[/tex], then we can factorize [tex]f[/tex] as

[tex]f(x) = x^2 + (k - 6) x + 9 = (x - \alpha) (x - \beta)[/tex]

Expand the right side and match up coefficients:

[tex]x^2 + (k-6) x + 9 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -(k-6) \\ \alpha \beta = 9 \end{cases}[/tex]

Now, recall that [tex](x+y)^2 = x^2 + 2xy + y^2[/tex]. It follows that

[tex]\boxed{\alpha^2 + \beta^2} = (\alpha + \beta)^2 - 2\alpha\beta = (-(k-6))^2 - 2\times9 = \boxed{k^2 - 12k + 18}[/tex]

and

[tex]\boxed{\alpha^2\beta^2} = 9^2 = \boxed{81}[/tex]

(b) If [tex]9(\alpha^2+\beta^2) = 2\alpha^2\beta^2[/tex], then

[tex]9 (k^2 - 12k + 18) = 2\times81 \implies 9k^2 - 108k = 0 \implies 9k (k - 12) = 0[/tex]

Since [tex]k\neq0[/tex], it follows that [tex]\boxed{k=12}[/tex].

(c) The simplest quadratic expression with roots [tex]\frac1{\alpha^2}[/tex] and [tex]\frac1{\beta^2}[/tex] is

[tex]\left(x - \dfrac1{\alpha^2}\right) \left(x - \dfrac1{\beta^2}\right)[/tex]

which expands to

[tex]x^2 - \left(\dfrac1{\alpha^2} + \dfrac1{\beta^2}\right) x + \dfrac1{\alpha^2\beta^2}[/tex]

Reusing the identity from (a-i) and the result from part (b), we have

[tex]\left(\dfrac1\alpha + \dfrac1\beta\right)^2 = \dfrac1{\alpha^2} + \dfrac2{\alpha\beta} + \dfrac1{\beta^2} \\\\ \implies \dfrac1{\alpha^2} + \dfrac1{\beta^2} = \left(\dfrac{\alpha + \beta}{\alpha\beta}\right)^2 - \dfrac2{\alpha\beta} = \left(\dfrac{-(k-6)}9\right)^2 - \dfrac29 = \dfrac29[/tex]

We also know from part (a-ii) that [tex]\alpha^2\beta^2=81[/tex].

So, the simplest quadratic that fits the description is

[tex]x^2 - \dfrac29 x + \dfrac1{81}[/tex]

To get one with integer coefficients, we multiply the whole expression by 81 to get [tex]\boxed{81x^2 - 18x + 1}[/tex].

The correct option is Option D: Yes, the graph passes the vertical line test.

The function is a relationship between two distinct sets X and set Y which can be many-one or one-one. here set X is called the domain and set Y is called the codomain.

The vertical line test states that

If we draw a straight vertical line( which is also parallel to the y-axis) and it touches the graph at only one point at all locations, then that relation is said to be a function and this relation will be also one-one.

So here in this function shown in the graph.

If we draw a vertical line parallel to the y-axis in this at any location then it crosses the graph only once. So, it passes vertical line test. And this graph is a function. Therefore option D is correct.

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Answer:

angle 1 is 105degrees and angle 2 is 75degrees

Step-by-step explanation:

The angle below angle 2, presumably angle 6, is also 75 degrees since line c and line d are parallel. If angle 2 is 75 degrees then we know that angle 1 is 105 degrees since both angles equal a straight line and a straight line has 180 degrees.

Step-by-step explanation:

Let X be the mail arrival time to a department that follows uniform distribution over 0 to 60 minutes.

The probability function of X is:

f

(

x

)

=

1

60

,

0

<

x

<

60

Now, the probability that the mail arrival time is more than 40 minutes on a given day is calculated below:

P

(

X

>

40

)

=

∫

60

40

1

60

d

x

=

[

x

60

]

60

40

Answer:

64 is the answer

Step-by-step explanation:

Mark me brainliest answer pls

Answer: A. 6, 18, 54, 162

Step-by-step explanation:

Each term is 3 times the last, meaning the common ratio is 3.

Substituting into point-slope form,

[tex]y+5=-\frac{5}{3}(x-15)\\\\y+5=-\frac{5}{3}x+25\\\\\boxed{y=-\frac{5}{3}x+20}[/tex]

Answer:

n = [tex]\frac{1}{36}[/tex]

Step-by-step explanation:

n × 18 = [tex]\frac{1}{2}[/tex]

To find the value of n we have to divide both sides by 18.

Let us solve it now.

18n = [tex]\frac{1}{2}[/tex]

[tex]\frac{18n}{18}[/tex] = [tex]\frac{1}{2}[/tex] ÷ 18

n =  [tex]\frac{1}{2}[/tex] ÷ 18

n = [tex]\frac{1}{2}[/tex] ÷ [tex]\frac{18}{1}[/tex]

n =  [tex]\frac{1}{2}[/tex] × [tex]\frac{1}{18}[/tex]

n = [tex]\frac{1}{36}[/tex]

Answer:

The volume of the pyramid with a base area of s² and the heights iss

The formula for calculating the volume of a square based pyramid is expressed as:

Volume = 1/3* Base area* Height

Since it is a square-based pyramid, the base area is calculated as:

Base Area = s ^ 2

If the height of the pyramid is equal to of the length of a side on the base, hence h = s

Substituting the base area and the height into the formula for the volume, this will give;

V = BH / 3

V = (s ^ 2 * s)/3

V = (s ^ 3)/3

V = 1/3 * s ^ 3

Hence the volume of the pyramid with a base area of s ^ 2 and the height s is 1/3 * s ^ 3

The range of y = sin(x) is [-1,1]

The function is given as:

y = sin(x)

The above function is the parent function of a sine function.

A sine function has a minimum of -1 and a maximum of 1.

This is represented by the interval [-1,1]

Hence, the range of y = sin(x) is [-1,1]

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∠1 and ∠2 are congruent because they are a pair of alternate interior angles , Option C is the right answer.

When two lines do not meet at any point and the distance between them remains constant , then the two lines are called parallel to each other .

It is given that the parallel lines are cut by a transversal and the angles are marked

The angle 1 and angle 2 are congruent because they are pair of alternate interior angles .

Therefore Option C is the right answer.

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Using proportions, considering the relation between the height and the shadow, it is found that the cell phone tower is 50 feet.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

When the shadow is of 8 feet, the height is of 10 feet. What is the height when the shadow is of 40 feet? The rule of three is:

8 feet - 10 feet

40 feet - h feet

Applying cross multiplication:

8h = 10 x 40

Simplifying by 8:

h = 10 x 5 = 50 feet.

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The average speed of sneeze is about 100 miles per hour.

The average speed of sneeze is determined from the total distance traveled by the sneeze to the total time of motion of the sneeze.

Averagely a sneeze can travel as fast as 100 miles in an hour, which is equivalent to 44.7 m/s.

Thus, the average speed of sneeze is about 100 miles per hour.

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Answer:

  (a)  5440.42 g

Step-by-step explanation:

The amount remaining (Q) is given in terms of the initial amount (Q₀) by the exponential decay formula ...

  Q = Q₀(1/2)^(t/330) . . . . . where t is in days

__

The amount after 540 days is ...

  1750 g = Q₀(1/2)^(540/330) = 0.321666Q₀

  Q₀ = (1750 g)/(0.321666) ≈ 5440.42 g

The initial quantity was about 5440.42 grams.

Answer:

Step-by-step explanation:

[tex]\sqrt{2} cos x-1=0\\cos x=\frac{1}{\sqrt{2} } =cos (\frac{\pi }{4} ),cos (2\pi -\frac{\pi }{4} )\\cos x=cos(2n\pi +\frac{\pi }{4} ),cos(2n\pi +\frac{7\pi }{4} )\\x=2n\pi +\frac{\pi }{4} ,2n\pi +\frac{7\pi }{4} \\n=0\\x=\frac{\pi }{4} ,\frac{7\pi }{4}[/tex]

Answer:

There is no line plot

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

[tex] \frac{9 - 1}{3 - 1} = 4[/tex]

The average rate of change on the interval [3, 9] will be 4.

The average rate of change between two points is given by the slope formula:

 m = average rate of change = (y2 -y1)/(x2 -x1)

The sequence an = 1(3)n − 1 is graphed below:

 m = (9 -1)/(3 -1) = 8/2

 m = 4

The average rate of change on the interval [3, 9] will be 4.

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1) [tex]V=lwh=5(x)(8)=\boxed{40x}[/tex]

2) [tex]V=\pi r^{2}h=(\pi)(3^{2})(8)=\boxed{72}\pi[/tex]

3) [tex]40x=72\pi\\\\x=\frac{72\pi}{40} \approx \boxed{5.7}[/tex]

[tex] \sf{\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve for x ~

[tex]\qquad \sf  \dashrightarrow \: 5x = 3x + 12[/tex]

[ by vertical opposite angle pair ]

[tex]\qquad \sf  \dashrightarrow \: 5x - 3x = 12[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2x = 12[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 12 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = 6[/tex]

Therefore, the correct choice is D. 6

Using translation concepts, the equation for the transformed logarithm is given by:

[tex]f(x) = \log{(x + 3)} - 3.48[/tex]

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

The logarithm function is given by:

[tex]f(x) = \log{x}[/tex].

We have that:

Hence:

[tex]f(x) = \log{x + 3} + b[/tex]

To find the shift up/down, we consider that f(0) = -3, hence:

[tex]-3 = \log{0 + 3} + b[/tex]

[tex]b = -3 - \log{3}[/tex]

b = -3.48.

Hence the function is given by:

[tex]f(x) = \log{x + 3} - 3.48[/tex]

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Answer:

The answer is A. 21

i hope this can help you! :)

Answer:

a)21

-step explanation:

[tex]\text{First I would recommend replacing the question mark with x}\\\text{We have two ratios, }\\\text{and we need to solve that hard-looking equation}\\\text{in terms of x}[/tex]

[tex]\rule{300}{1.7}[/tex]

[tex]\text{We can multiply x times 36:: 36x}\\\text{We can multiply 4 times 27:: 108}[/tex]

[tex]\rule{300}{1.7}[/tex]

[tex]\text{Now it's easier to solve for x:: 36x=108; x=3}\\\text{ (we divided by 36 on both sides of the = sign)}[/tex]

[tex]\rule{300}{1.7}[/tex]

[tex]\text{The missing number is 3}[/tex]

Answer:

it is 18 by unknown number

[tex]A \cup B[/tex] represents the set that contains all the elements that are in at least one of the sets, so [tex]A \cup B=\{1, 2, 5, 6, 7, 8, 9, 10, 12, 16, 18, 19, 20, 21, 22, 23, 25\}[/tex].

We want the complement of this set (in other words, the set with all the elements in the universal set but not in the given set).

This set is {3, 4, 11, 13, 14, 15, 17, 24}

Answer:

See below

Step-by-step explanation:

She combined 3x and - 9x   incorrectly

  she should have gotten  - 6 x       not  - 12 x

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:2x - 3 [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: (f - g)(x)[/tex]

[tex]\qquad \tt \rightarrow \: f(x) - g(x)[/tex]

Simple procedure :

[tex]\qquad \tt \rightarrow \: 3x - 1 - (x + 2)[/tex]

[tex]\qquad \tt \rightarrow \: 3x - 1 - x - 2[/tex]

[tex]\qquad \tt \rightarrow \: 3x - x - 1 - 2[/tex]

[tex]\qquad \tt \rightarrow \: 2x - 3[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞