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kari74 [83]
2 years ago
15

Which point is best represented by the ordered pair (2, -5.5)?

Mathematics
2 answers:
Llana [10]2 years ago
7 0

Answer:

B

Step-by-step explanation:

Delvig [45]2 years ago
4 0

Answer:

The answer is A, Point W.

This is because (2, -5.5) is equal to (x, y). So, x would be 2, while point y will be -5.5. Point W represents this because it goes to 2 on the x axis, and down to -5.5 on the y axis.

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What is the distance between (3,2) and (3,-8)
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The answer is (0,-10)
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3 years ago
Solve the equation y = ax - b for the variable a.
lorasvet [3.4K]

Answer:

y-b/x = a

Step-by-step explanation:

y-b/x = a

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3 years ago
The data set represents the total number of tickets each person purchased for a play.
Bezzdna [24]

Median: The middle number when all the numbers are listed in order.

First, we would put our numbers in order from least to greatest.

0 - 0 - 1 - 1 - 1 - 2 - 2 - 2 - 4 - 4

Next, we need to find the middle number. When we cross of one number from the left and one number from the right and keep doing this, we come across two numbers that are in the middle. The two numbers that are in the middle are 1 and 2. We find the median by adding 2 and 1 together to get a sum of 3 and then divide it by 2 to get an answer of 1.5

The median of this set of numbers is 1.5

4 0
3 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

7 0
3 years ago
Find the missing side lengths. Leave your answers as radicals in the simplest form.
maks197457 [2]

Answer:

Does the answer help you?

4 0
3 years ago
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