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3 years ago

The figure above the x-axis is the preimage. The figure below the x-axis is the image. Which transformation produced the image?

Mathematics

1 answer:

Sav [38]3 years ago

5 0

Answer:

Transformation : $(x,y)\rightarrow (x,-y)$

Step-by-step explanation:

We are given pre-image is above x-axis abd original figure below x-axis.

As we know y-axis below x-axis always negative. When we translate about x-axis only y value of each coordinate change.

x-coordinates remain same.

Best transformation for below x-axis to above x-axis would be $(x,y)\rightarrow (x,-y)$ .

For example, (2,-3) change to (2,3)

This is valid to all types of figures which form below x-axis.

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Please help me out with this :)

yan [13]

Answer:

y = x - 5

Step-by-step explanation:

Given the equation

y = x + ?

We are being asked what value is added to x to give y

Consider the table, that is

x = 1 → y = - 4

x = 2 → y = - 3

x = 3 → y = - 2

x = 4 → y = - 1

x = 5 → y = 0

x = 6 → y = 1

In each case 5 is being subtracted from x to obtain y, that is

y = x - 5 ← equation relating x and y

6 0

3 years ago

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Mrrafil [7]

Fue answer is poseitive3

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3 years ago

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Sam buys tickets to a basketball game for $92.00 total. He must also pay 11% sales tax on the tickets. Write and simply a numeri

Flauer [41]

Sales tax 11%
11/100 x $92.00
=$10.12

Total cost = x
x= $92.00 + $ 10.12
= $102.12

7 0

4 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

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3 years ago

Please Simplify 60/105 to lowest form

svlad2 [7]

4/7. Divide the numerator and denominator by 15.

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3 years ago