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3 years ago

This diagram shows a pre-image △ABC , and its image, △A′′B′′C′′ , after a series of transformations.

Mathematics

2 answers:

Stella [2.4K]3 years ago

8 0

First ABC is reflected across the y axis to become A'B'C'

Then A'B'C' is rotated 90 counterclockwise about the origin to make it A"B"C"

Because the transformations are both rigid, the pre-image and image are congruent

Hope this helped

sladkih [1.3K]3 years ago

7 0

Answer:

Step-by-step explanation:

ΔABC is reflected across y axis [ as we can see points A',B,'C' and A,B,C are equidistant from y - axis. y- axis is working like a mirror ]

ΔA'B'C' is rotated 90 degrees counterclockwise about the origin to become A"B"C". [Sides A'B' and A"B" are perpendicular to each other]

Because the transformations are both rigid [No dilation in the figure A"B"C"], the pre image and the image are congruent [All corresponding sides are equal].

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State whether each sequence is arithmetic and justify your answer. If the sequence is arithmetic, write a recursive and an expli

nasty-shy [4]

Answer:

Part A

$f(n)=52-12(n-1)$

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$ Geometric Sequence

Part C

1/4,3/4,5/4,7/4,9/4

$g(n)=\frac{1}{4}+\frac{2}{4}(n-1)\\f(n)=\left\{\begin{matrix}1/4if \: \:n=1 & \\ f(n+1)+2/4& if\: n\geq 2 \end{matrix}\right$

Part D:

$h(n)=1.1+0.4(n-1)\\h(n)=\left\{\begin{matrix}1.1 & if\:n=1 \\ h(n+1)+0.4 & if\:n\geq 2\end{matrix}\right$

Step-by-step explanation:

By definition, an Arithmetic Sequence holds the same difference between each following number.

Part A

$(52,40, 28, 16)\\52-40=12\\40-28=12\\28-16=12\\d=12$

Explicit Formula

To write an explicit formula is to write it as function.

$f(n)=52-12(n-1)$

Recursive Formula

To write it as recursive formula, is to write it as recurrence given to some restrictions:

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$

Geometric Sequence, since 2*2=4 8*2=16 and 16*2=32 and 8+2=10 8+16=24

Part C

$(\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4})\\\$

Arithmetic Sequence, difference

$d=\frac{2}{4}$

Explicit Formula:

$g(n)=\frac{1}{4}+\frac{2}{4}(n-1)$

Recursive Formula

$g(n)=\left\{\begin{matrix}\frac{1}{4} &if\:n=1 \\ g(n+1)+\frac{2}{4} &if\: n\geq 2\end{matrix}\right.$

Part D

(1.1,1.5,1.9,2.3,2.7) Arithmetic Sequence, difference d=0.4

Explicit formula

$h(n)=1.1+0.4(n-1)\\$

Recursive Formula

$h(n)=\left\{\begin{matrix}1.1 &if\:n=1 \\ h(n+1)+0.4 &if\: n\geq 2\end{matrix}\right.$

6 0

3 years ago

jim is planning a party. he spends $150 on a room for the party. he also spends $16.50 per person for food. jim finds the total

meriva

$577.50is the amount he paid.

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2 years ago

Without using a calculator, fill in the blanks with two consecutive integers to complete the following inequality ____ < √134

puteri [66]

$11 < \sqrt{134} < 12\\\\because\\\\11^2=121 < (\sqrt{134})^2=134 < 12^2=144$

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2 years ago

Read 2 more answers

What is the value of x if x/4+2=15 Help!

NNADVOKAT [17]

Answer:

Step-by-step explanation:

because i don't know if the 2 is with the 4 or not I have calculated both.

x/4 + 2 = 15

<=> x/4 = 13

<=> x = 13.4 = 52

x/(4+2) = 15

<=> x/6 = 15

<=> x = 6.15 = 90

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2 years ago

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14 friends go out for dinner and the bill for food and drink comes to £130. The friends decided to add a 5% tip on the top of th

Hitman42 [59]

The amount each friend does pay is £9.75

<h3>How to calculate the pay of each friend using fractions and arithmetic operations?</h3>

We are going to find 5% of the total amount paid and divide it by the numbers of all the friends to determine the amount paid by each one of them.

From the given information:

The 5% of 130 = £6.5
Total amount paid = £130 + £6.5 = £136.5

Therefore, the cost of each friend is = $\dfrac{136.5}{14}$

The cost of each friend = £9.75

Learn more about fractions here:

brainly.com/question/78672

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1 year ago

This diagram shows a pre-image △ABC , and its image, ​ △A′′B′′C′′ ​ , after a series of transformations.

This diagram shows a pre-image △ABC , and its image, △A′′B′′C′′ , after a series of transformations.