Ask question

Ask question

All categories

3 years ago

12

3/2 x 3 x 1/5 please help me

1 answer:

USPshnik [31]3 years ago

5 0

Answer:

9/10 is the answer.

You might be interested in

What are the factors of -90?

Helen [10]

-90 -45 -30 -18 -15 -10 -6 -5 -3 -2 -1

8 0

3 years ago

Read 2 more answers

Help!!!!!!!!!!!!!!!!!!!!

yaroslaw [1]

<h2>Answer:</h2><h2>x=3 </h2><h2>p=3csc (β)sec(θ)</h2><h2>hope this helps! ;)</h2>

6 0

2 years ago

Let f(x) = 4x – 5 and g(x) = 3x + 7. find f(x) + g(x) and state its domain.

Tema [17]

We are given the functions:

f(x) = 4 x – 5 ---> 1

g(x) = 3 x + 7 ---> 2

To find for the value of f(x) + g(x), all we have to do is to add equations 1 and 2:

f(x) + g(x) = 4 x – 5 + 3 x + 7

f(x) + g(x) = 7 x + 2 = y

In this case, for any real number value assign to x, we get a real number value of y. This is because the function is linear.

Therefore the domain of the function is all real numbers.

5 0

3 years ago

Suppose that theta is an angle in standard position whose terminal side Intersects the unit circle at (-11/61, -60/61)

Blizzard [7]

Answer:

The exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

Step-by-step explanation:

The components of the unit vector are $x = -\frac{11}{61}$ and $y = -\frac{60}{61}$ . Since $r = 1$ , then $x = \cos \theta$ and $y = \sin \theta$ . By Trigonometry, tangent and secant can be calculated by the following expressions:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{x}$

Now, the exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

3 0

3 years ago

A square and a circle intersect so that each side of the square contains a chord of the circle equal in length to the radius of

krek1111 [17]

Answer:

Step-by-step explanation:

it is given that Square contains a chord of of the circle equal to the radius thus from diagram

$QR=chord =radius =R$

If Chord is equal to radius then triangle PQR is an equilateral Triangle

Thus $QO=\frac{R}{2}=RO$

In triangle PQO applying Pythagoras theorem

$(PQ)^2=(PO)^2+(QO)^2$

$PO=\sqrt{(PQ)^2-(QO)^2}$

$PO=\sqrt{R^2-\frac{R^2}{4}}$

$PO=\frac{\sqrt{3}}{2}R$

Thus length of Side of square $=2PO=\sqrt{3}R$

Area of square $=(\sqrt{3}R)^2=3R^2$

Area of Circle $=\pi R^2$

Ratio of square to the circle $=\frac{3R^2}{\pi R^2}=\frac{3}{\pi }$

5 0

3 years ago