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2 years ago

Find the length of the other two side of the isosceles right triangle

Mathematics

1 answer:

Valentin [98]2 years ago

4 0

Answer:

The other two sides are 7 units each.

Step-by-step explanation:

The triangle is isosceles.

The sides are in the ratio of

$x : x : x \sqrt{2}$

The hypotenuse is

$7\sqrt{2}=x \sqrt{2}\\\\x = 7$

So, the other two sides are 7 units each.

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A rectangle has length X +5 and width X +4 if the perimeter of the rectangle is 42 m what are the length and width of the rectan

Maru [420]

The perimeter is 2(length) + 2(width), so 2(x+5) + 2(x+4) = 42 m. In solving for x, you find 6. Plug 6 back into your length and width where x originally was, and you find your length to be 11m, and your width to be 10m. Hope this helps!

6 0

3 years ago

Read 2 more answers

While in France, Judy wants to sign up for a bike tour to see the countryside. She likes a tour that is 40 kilometers long, but

Sonja [21]

Answer:

24 miles

Step-by-step explanation:

In order to calculate Judy's estimation, we would simply have to multiply the actual distance in kilometers of the tour by the number of miles that Judy believes are in a single mile. This would give us Judy's estimation for how long the tour would be in miles.

40 km. * 0.6 miles = 24 miles

Finally, we can see that Judy's estimation would be that the tour is 24 miles long. Using Judy's believed conversion rate.

8 0

2 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

<u>Trigonometric Identities</u>

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$