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3 years ago

How many kilometers did Amelia drive per liter

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

8 kilometers

hope it helps

Step-by-step explanation:

6 liters = 48 kilometers

48 / 6 = 8

1 liter = 8 kilometers

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Which angle measures are correct? Select three options,

zhenek [66]

Answer:

m∠W = 125°

m∠X = 55°

m∠Z = 55°

Step-by-step explanation:

The explanation of angle measures is shown below:-

The shape which is parallelogram WXYZ. m∠Y = 125

A parallelogram represents the shape of quadrilateral in which both opposite sides and opposite angles are similar to each other. One parallelogram diagonals bisect each other. Consecutive angles are also supplementary to a parallelogram.

In WXYZ parallelogram is there

m∠Z + m∠Y = 180° is an angle of Consecutive in a parallelogram is supplementary

m ∠ Z + 125 = 180

m ∠ Z = 180 - 125

m ∠ Z = 55°

m ∠ W which equals to m∠Y where equal angles are Opposite

m∠W = 125°

m ∠ W + m ∠ X = 180°

which is the angle of Consecutive in a parallelogram called supplementary)

m ∠ X + 125 = 180

m ∠ X = 180 - 125

After solving the above equation we will get

m ∠ X = 55°

8 0

3 years ago

Given triangle ABC, which equation could be used to find the measure of ∠B? right triangle ABC with AB measuring 6, AC measuring

rjkz [21]

Answer:

$cos B = \dfrac{2\times \sqrt5}{5}$ is the correct answer.

Step-by-step explanation:

We are given a right angled $\triangle ABC$ with the side measurements as:

AB = 6

AC = 3

BC = 3 $\sqrt5$

In a right angled triangle, the angle of $90^\circ$ is the largest angle and the side opposite to the right angle is the largest.

We are given that the side BC = 3 $\sqrt5$ which has a value greater than 6 which means side BC is the largest side.

And angle opposite to BC is $\angle A$ is the largest i.e.

$\angle A=90^\circ$

Please refer to the attached image for the given dimensions of the triangle.

Now, we can apply trigonometric rules to easily find out the value of $\angle B$

$cos \theta = \dfrac{Base}{Hypotenuse}\\OR\\cos B = \dfrac{AB}{BC}\\\Rightarrow cos B = \dfrac{6}{3\sqrt5}\\\Rightarrow cos B = \dfrac{2}{\sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{\sqrt5 \times \sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{5}$