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3 years ago

An angle in standard position measures -7pi/6 radians.

Mathematics

1 answer:

sukhopar [10]3 years ago

3 0

Answer:

Quadrant II

Step-by-step explanation:

We can convert $-\frac{7\pi}{6}$ radians to degrees by multiplying by $\frac{180}{\pi}$ to get $-210^{\circ}$ .

$-210^{\circ}$ means we are rotating $210^{\circ}$ clockwise, so you would end up in Quadrant II.

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A bag contains 4 black tiles, 5 white tiles, and 6 blue tiles. Event A is

Murljashka [212]

Answer:

2/21

Step-by-step explanation:

The question is badly formatted so cannot match the answer choice

Total number of tiles = 4 + 5 + 6 = 15

P(A) = P(drawing a black tile from 15 tiles) = 4/15

After event A, there are 14 tiles left : 3 black, + 5 white + 6 blue

P(B/A) = P(drawing white tile after a black tile has been drawn) = 5/14

P(A and B) = P(A).P(B/A) = 4/15 x 5/14 = 2/21

Note that P(B) = 5/15 since we are talking about independently drawing a white tile from the original stack of tiles

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago

Three and 2/3÷1 and 3/11

love history [14]

Answer:

3.93939394

rounded it would be 3.934

Step-by-step explanation:

5 0

3 years ago

You have two exponential functions. One function has the formula g(x) = 5 x . The other function has the formula h(x) = 5-x . Wh

Marrrta [24]

Hmm it would be k(x)=2(5x) or it could be k(x)=10x

Hope this helps!

5 0

3 years ago

Read 2 more answers