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N76 [4]
3 years ago
7

Whenever she visits Oakdale, Karin has to drive 55 kilometres due north from home. Whenever she visits Brookfield, she has to dr

ive 48 kilometres due east from home. How far apart are Oakdale and Brookfield, measured in a straight line?
Please show work <3

Mathematics
2 answers:
Natalija [7]3 years ago
6 0
Sorry it is a bit messy but the answer is right

inessss [21]3 years ago
4 0
Here it is, just use pitagorean theorem, because it's a right triangle

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its 16 cm, so 2 full rotations would be 16 times 2

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Given the values in the table on the left come from
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–3.2 < x < –2.4

1.6 < x < 2.4

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What is 2/50 as a decimal
Thepotemich [5.8K]
Okay so what we need to do is simplify 2/50, which is 1/25.
Now divide 25 into 1.
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Which steps transform the graph of
vitfil [10]

Answer:

The correct option is the last one.

Step-by-step explanation:

To transform the graph of y = x ^ 2 into y = -2(x - 2) ^ 2 + 2 the following steps are fulfilled:

1) Move the graph 2 units to the right:

Let y = f (x-2) then y =(x-2) ^ 2 Notice that the cut point has been moved to x = 2.

2) Reflect on the x axis:

To reflect a graph on the x-axis we do y = -f(x) Then f (x) = -(x-2) ^ 2

3) Stretch according to factor 2.

For this we do y = 2f(x)

Then we have f(x) = -2 (x-2) ^ 2

4) Move up the graph in two units:

We do y = f(x) +2

Then y = -2(x-2) ^ 2 +2.

These steps coincide with those listed in the last option. Therefore the correct option is the last one.

"Translate 2 units on the right, reflect on the x-axis, stretch according to the factor 2 and translate 2 units"

4 0
3 years ago
In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. The spike driv
Fynjy0 [20]

Answer:

4.75 seconds

Step-by-step explanation:

Initial height of the ball from the ground = 10 feet

Upward initial velocity given = 55 feet per second

Value of g ( acceleration due to gravity) = 32.2 feet per s²

Motion of the ball:

The ball first goes vertically upwards, gravity decelerates the body and it momentarily comes to rest at a point in its upward trajectory, which is the point of maximum height. From this point, to the ground, the ball behaves as a freely dropped body.

Till the ball reaches its maximum height:

u = + 55

v = 0 ( final velocity is zero)

a = - 32.2 (since it is deceleration)

we know that <em>v = u + at </em>

⇒    0 = 55 - 32.2t

⇒  t = 1.7 s

Also , we have <em>s = ut + (1/2)at²</em>

Here, s is the maximum height from the point where ball is thrown

So,     s = 55(1.7) - (0.5)(32.2)(1.7)(1.7)

⇒ s = 140 feet

So at a height of 140 feet + 10 feet (initial height) = 150 feet, the ball acts as a freely dropped body.

Here,      u = 0

               a = +32.2

               s= 150

<em>s = ut + (1/2)at²</em>

⇒  150 = 0.5 ( 32.2) (t²)

⇒  t² = 300/32.2 = 9.31

⇒ t = 3.05 sec

So total time = 1.7 + 3.05 = 4.75 seconds

8 0
3 years ago
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