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Aleksandr-060686 [28]
4 years ago
10

Which statement is true about point Y?

Mathematics
1 answer:
Viefleur [7K]4 years ago
3 0

Answer:C

Step-by-step explanation:

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0.125 into a decimal
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It’s already a decimal
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3 years ago
F(x)=x+7 g(x)=4x^2-9 h(x)=-2x+1 what is the expression that represents f(x)+h(x)
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f(x) = x + 7

h(x) =  - 2x + 1

Add them up :

f(x) + h(x) = x + 7  - 2x + 1 \\

Collect like terms

f(x) + h(x) =  - x + 8

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3 years ago
Calvin took his friends to a movie for his birthday. The movie tickets cost $12 per person. He spent less than $180.If x is the
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The inequality should be
12 \leqslant 12x  < 180
The sign on the right is only a 'smaller than' sign because he spent less than 180.
On the left, assuming that even if Calvin took nobody (poor Calvin), we should assume that he at least went himself, hence he spent a minimum of $12.
4 0
3 years ago
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Eric ran from school to the town monument and back again. On his way to the monument, he ran at 10kph and went back to school at
Aloiza [94]

Answer:

Distance = 10 km

Step-by-step explanation:

Let x be the number of hours taken from school to town and y be the number of hours taken back to the school

Then distance covered during first trip would be 10x (distance = speed*time) and during the second trip would be 8y. Both distances are equal.

=> 10x = 8y  

Dividing both sides by 2

=> 5x = 4y

=> 5x-4y = 0   ------------------(1)

<u><em>The total time for both the trips is:</em></u>

=> x + y = 2.25 -------------------(2)

Multiplying eq (2) by 5

=> 5x+5y = 11.25  ---------------(3)

Subtacting (3) from (1)

=> 5x-4y-5x-5y = 0-11.25

=> -4y-5y = -11.25

=> -9y = -11.25

Dividing both sides by -9

=> y = 1.25 hrs

Putting in (2)

=> x + 1.25 = 2.25

=> x = 2.25 - 1.25

=> x = 1 hr

<u><em>Now, Calculating the Distance</em></u>

=> Distance = 10x

=> 10 ( 1 )

=> Distance = 10 km

3 0
4 years ago
1.<br> How can you count following distance?
anyanavicka [17]

At a MINIMUM, during dry weather conditions, you should have at least 2 seconds of space between you and the vehicle in front of you (3 seconds is better). Do this by using a fixed object such as a bridge, tree, or even a crack or shadow in the roadway.

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