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garik1379 [7]
3 years ago
7

Point P(2, 3) is reflected across the y-axis to create point P'. Determine the quadrant of the image.

Mathematics
2 answers:
Gnoma [55]3 years ago
7 0
The reflection of P(2, 3)  across the y-axis has the coordinate P'(-2,3)

& it is located in the 2nd quadrant
Anarel [89]3 years ago
4 0

Answer:

2nd quadrant.

Step-by-step explanation:

The given point is P(2,3). This is in 1st quadrant.

Now, this point is reflected across the y-axis.

When we reflect this point p about the y-axis then the y -coordinate remains same but the x- coordinate will be the negative of previous one.

Therefore, the coordinate of the point after reflection is P'(-2,3)

Here x coordinate is negative and y coordinate is positive.

Hence, the point P'(-2,3) is in 2nd quadrant.

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If a fair die is rolled 6 times, what is the probability, rounded to the nearest
trapecia [35]

The probability of rolling the die once to get a four is 1/6

So we do 1/6*1/6*1/6 which is 1/216 since it's combined probabilities

Hope this helped

8 0
2 years ago
AABC is similar to AAXY by a ratio of 4:3. If BC = 24, what is the length of XY?
Alenkasestr [34]

Answer:

OXY=18

Step-by-step explanation:

The ratio of AABC to AAXY is 4:3 so the lengths will be in that ratio too. The ratio of BC:XY is 4:3. BC is 24 so 24:XY is 4:3. You need to divide 24 by 6 to get 4 so multiply 3 by 6 to get the answer 18.

7 0
3 years ago
Plz someone help on this question​
sergey [27]

Answer:

Hi there!

Your answer is:

1050 liters per 6 days

Step-by-step explanation:

350liters per 2 days

*3 to determine 6 days

1050 liters per 6 days

Hope this helps

5 0
3 years ago
Read 2 more answers
Hi, I'm struggling on level 2 for
andrew11 [14]

Answer:

Step-by-step explanation:

we know 100 cm are in one meter,  so,  3 meters = 300 cm

then  20/300  =  20 cm  of 3 meters  , now just reduce the fraction

20/300

= 2 / 30

= 1/15

see??

6 0
2 years ago
A ball dropped from the top of a building can be modeled by the function f(t)=-16^2+36, where t represents time in seconds
Viefleur [7K]

Answer:

The ball is dropped from a height of 36 feet

The bee and the ball will collide after approximately 1.3 seconds

The bee and the ball will collide at approximately 8 feet above the ground

The ball hits the ground after 1.5 seconds

Step-by-step explanation:

<u><em>The complete question is</em></u>

A ball dropped from the top of the building can be modeled by the function f(t)=-16t^2 + 36 , where t represents time in seconds after the ball was dropped. A bee's flight can be modeled by the function, g(t)=3t+4, where t represents time in seconds after the bee starts the flight.

The graph represents the situation.

select all that apply

1) The bee launches into flight from the ground.

2) The ball is dropped from a height of 36 feet.

3) The bee and the ball will collide after approximately 1.3 seconds.

4) The bee and the ball will collide after approximately 8 seconds.

5) The bee and the ball will collide at approximately 8 feet above the ground.

6) The bee and the ball will not collide.

7) The ball hits the ground after 1.5 seconds

see the attached figure to better understand the problem

<u><em>Verify all the options</em></u>

case 1) The bee launches into flight from the ground.

The statement is false

Because

we have

g(t)=3t+4

For t=0

g(t)=3(0)+4=4\ ft

That means ----> The bee launches into flight from a height 4 feet above the ground

case 2) The ball is dropped from a height of 36 feet

The statement is true

Because

For t=0

f(t)=-16(0)^2+36=36\ ft

case 3) The bee and the ball will collide after approximately 1.3 seconds

The statement is true

Because

Equate f(t) and g(t)

-16t^2+36=3t+4

-16t^2-3t+32=0

solve the quadratic equation by graphing

The solution is t=1.324 sec

see the attached figure N 2

case 4) The bee and the ball will collide after approximately 8 seconds

The statement is false

Because, the bee and the ball will collide after approximately 1.3 seconds (see case 3)

case 5) The bee and the ball will collide at approximately 8 feet above the ground

The statement is true

Because

For t=1.324 sec

substitute the value of t in f(t) or g(t)

g(t)=3(1.324)+4=7.97\ ft

case 6) The bee and the ball will not collide

The statement is false

see case 3) and case 5)

case 7) The ball hits the ground after 1.5 seconds

The statement is true

Because

we know that

The ball hit the ground when the value of f(t) is equal to zero

so

For f(t)=0

-16t^2+36=0

t^2=\frac{36}{16}

t=\frac{6}{4}=1.5\ sec

4 0
3 years ago
Read 2 more answers
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