All categories

3 years ago

What is the height of this triangle between C and D

Mathematics

1 answer:

Rom4ik [11]3 years ago

8 0

Answer:

12 units

Step-by-step explanation:

To find any side length of a right triangle given two side lengths, you would use the pythagorean theorem which is a² + b² = c². In this equation, a and b represents either the height of base of the triangle, and c represents the hypotenuse of the triangle (the diagonal line - in your question it is 13). By plugging in the base and hypotenuse, you get the equation of 5² + b² = 13². 5² is equal to 25 and 13² is equal to 169, so the equation is 25 + b² = 169. Subtract 25 for both sides of the equation and you find that b² = 144. Square both sides and the missing length is 12 because √144 = 12.

You might be interested in

the cost to rent skis at a local sporting goods store is $15 plus $20 per day. which equation models the relationship between th

Andrew [12]

Answer:

c = 20d + 15

Step-by-step explanation:

answer choices may help if the answer is not one of the choices; but here is the answer that it should be:

20 dollars per day is 20 times the length of rental.

15 is a cost that is added and does not change.

c = the total cost therefore:

c = 20d + 15

4 0

3 years ago

Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

6 0

2 years ago

HHHHHHEEEEEELLLLLLPPPPPPP!!!!!!!

Klio2033 [76]

Check the picture below

if that red segment, GJ, is parallel to the AE base segment of the triangle, then, the segment GJ is the midsegment of the triangle, and by the side-splitter theorem, those two triangles are similar.