All categories

2 years ago

Analyze the reflectional symmetry of the partial circle.

Mathematics

2 answers:

finlep [7]2 years ago

8 0

1 because there is only one way you could “fold it”

Verizon [17]2 years ago

7 0

Answer:

i think its 2, but if im wrong im sorry

You might be interested in

If A=60 and B=60, then find the value of tanA(A+B) = tanA + tanA/1- tanA .tanB

mihalych1998 [28]

Answer:

your answer is 160

Step-by-step explanation:

just add 60+60_160 u might have to add because its tanned

6 0

2 years ago

There are 64 students in a speech contest. Yesterday, Ā of them gave their speeches. Today, 7 of the remaining students gave the

Alexxx [7]

Answer:

Step-by-step explanation:

Ā means the average

So it means the average of 64 gave their speech which should be 64/2 = 32

So 32 gave their speeches yesterday,

7 gave today

Total number that has given is

32 + 7 = 39

Therefore the number remaining are

62 - 39 = 23

8 0

2 years ago

Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

AVprozaik [17]

Answer:

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2<em>x</em>).

Note that since <em>x</em> is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(<em>x</em>) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since <em>x</em> is in QII, cosine and tangent will be negative while sine will be positive. Hence: