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lions [1.4K]
3 years ago
8

Point R is at (2, 1.2) and Point T is at (2, 2.5) on a coordinate grid. The distance between the two points is ____. (Input numb

ers and decimal point only, such as 8.2.)
Mathematics
1 answer:
Nataly [62]3 years ago
3 0
You could graph this, but it is simpler in this case to do it mathematically. 

Since the x and values are the same (implying undefined slope) you can subtract the smaller y value from the larger y value. 2.5-1.2=1.3. 
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Due rn!!! please help
Alex Ar [27]

Answer: The answer is A.

Step-by-step explanation:

4 0
2 years ago
Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking tw
kow [346]

Answer:

\frac{5}{13}

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

Since, here replacement is not allowed,

Thus, the probability of getting a green marble and a red marble

= first red and second green + first green second red

=\frac{3}{13}\times \frac{10}{12}+\frac{10}{13}\times \frac{3}{12}

=2\times \frac{3}{13}\times \frac{10}{12}

=\frac{30}{78}

=\frac{5}{13}

Note : The probability of getting a green marble first and a red marble second

= \frac{10}{13}\times \frac{3}{12}

=\frac{30}{156}

=\frac{5}{26}

4 0
3 years ago
Help me and ill mark branilest for first person to answer CORRECTLY. A histogram is being made for the following list of data. I
kolezko [41]
So to put them in intervals of 5 you would have:
70-74
75-79
80-84
85-89
90-94
95-99
In these intervals you would have:
70-74: 72
75-79: 78
80-84: 81
85-89: 86, 86, 87
90-94: 92, 92, 92
95-99: 98

The second interval is 75-79 where there is only one number. Therefore the frequency will be 1.
The correct answer is A. Hope this helps! :)
5 0
2 years ago
T/F I can take the square root of a negative number
Anvisha [2.4K]
1).  F
2).  T
3).  F
4).  F
5).  F

Bet you learned a lot from this answer !
6 0
3 years ago
In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the
emmasim [6.3K]

Answer:

a=\frac{2S -2v_ot-2s_o}{t^2}

Step-by-step explanation:

We have the equation of the position of the object

S = \frac{1}{2}at ^2 + v_ot+s_o

We need to solve the equation for the variable a

S = \frac{1}{2}at ^2 + v_ot+s_o

Subtract s_0 and v_0t on both sides of the equality

S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o

S -v_ot-s_o = \frac{1}{2}at ^2

multiply by 2 on both sides of equality

2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2

2S -2v_ot-2s_o =at ^2

Divide between t ^ 2 on both sides of the equation

\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}

Finally

a=\frac{2S -2v_ot-2s_o}{t^2}

5 0
3 years ago
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