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3 years ago

a car travels 30 1/5 miles in 2/5 of an a hour. what if the average speed, in miles, per hour of the car?

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

8 0

Answer:

Average speed of the car is $75.5\text{ miles per hour}$ .

Step-by-step explanation:

We are given the following information:

Distance traveled by the car =

$30\displaystyle\frac{1}{5}\text{ miles}$

Time taken by the car =

$\displaystyle\frac{2}{5}\text{ hours}$

Formula:

Speed = $\displaystyle\frac{\text{Distance covered}}{\text{Time taken}}$

Putting all the values,

$\text{Speed} = \displaystyle\frac{30\frac{1}{5}}{\frac{2}{5}} = \frac{\frac{151}{5}}{\frac{2}{5}} = \frac{151}{2} = 75.5\text{ miles per hour}$

Average speed of the car is $75.5\text{ miles per hour}$ .

alekssr [168]3 years ago

6 0

Answer:

Average speed = 30.2 miles / .4 hours =

75.5 mph

Step-by-step explanation:

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3 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$