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3 years ago

A stunt car drives off a 150m cliff at a speed of 45mph

Mathematics

1 answer:

seraphim [82]3 years ago

6 0

Answer:

5.53 seconds.

Step-by-step explanation:

The equation of motion is

d = 4.9t^2 so

150 = .9 t^2

t^2 = 150 / 4.9

t^2 = 30.612

t = 5.53 seconds.

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The velocity v of an object rolled down an inclined plane is given by the formula v=

drek231 [11]

Answer:

d = v/53

Step-by-step explanation:

Divide both sides by 53 to make d the subject of the formula

7 0

3 years ago

What is the missing number in the sequence shown? 5, 9, 14, 20, __, 35

Sonja [21]

$\sf{ 27}$

Step-by-step explanation:

5+4=9
9+5=14
14+6=20
20+7=27
27+8=35

So The sequence is 5,9,14,20,27,35

8 0

3 years ago

Read 2 more answers

Find all missing angles in the diagram shown. 4 کر) 47° 2 5 522

faltersainse [42]

I don’t see a diagram ? Just numbers...

5 0

3 years ago

Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if

Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}$

$JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}$

$JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}$

$JK=\sqrt{4+9}$

$JK=\sqrt{13}$

Similarly,

$KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6$

$LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}$

$JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6$

Now, perimeter P of ▱JKLM is

$P=JK+KL+LM+JM$

$P=\sqrt{13}+6+\sqrt{13}+6$

$P=2\sqrt{13}+12$

$P=2(3.61)+12$

$P=7.22+12$

$P=19.22$

$P\approx 19.2$

Therefore, the perimeter P of ▱JKLM is 19.2 units.

3 0

3 years ago

Points P, Q and R are collinear such that PQ : QR=2:3 point P is located at (1,3) and point R is located at (11,15)

balandron [24]

Using line segments, it is found that:

The coordinates of Q are (5,7.8).

The midpoint of segment PQ is M(6,9).

----------------------------

Point P is located at (1,3).
Point R is located at (11,15).
Point Q is located at (x,y).
PQ:QR = 2:3, which means that:

$Q - P = \frac{2}{5}(R - P)$

This is used to find the x and y coordinates of Q.

----------------------------

The x-coordinate of P is 1.
The x-coordinate of R is 11.
The x-coordinate of Q is x.

Thus:

$Q - P = \frac{2}{5}(R - P)$

$x - 1 = \frac{2}{5}(11 - 1)$

$x - 1 = \frac{2}{5}10$

$x - 1 = 4$

$x = 5$

----------------------------

The y-coordinate of P is 3.
The y-coordinate of R is 15.
The y-coordinate of Q is y.

Thus:

$Q - P = \frac{2}{5}(R - P)$

$y - 3 = \frac{2}{5}(15 - 3)$

$y - 3 = \frac{24}{5}$

$y - 3 = 4.8$

$y = 7.8$

The coordinates of Q are (5,7.8).

----------------------------

The midpoint of segment PQ is the mean of the coordinates, thus:

$M = (\frac{1 + 11}{2}, \frac{3 + 15}{2}) = (\frac{12}{2}, \frac{18}{2}) = (6,9)$

The midpoint of segment PQ is M(6,9).

A similar problem is given at brainly.com/question/24148182

4 0

3 years ago