Please help ASAP!

Mathematics

1 answer:

Misha Larkins [42]3 years ago

6 0

The probability of pulling a boys name first is 8/13. Then the probably of pulling a girls second is 2/5.

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Two sides of a triangle measure 5 in. and 12 in. Which could be the length of the third side?

Airida [17]

Answer: 10.9

Step-by-step explanation:

You would use Pythagorean theorem:

5x5+12x12=c

5x5=25

12x12=144

Now, 25+144=169

Now find the square root of 169 which is 13 which isn't an answer choice, which means your prolly after a leg.

I assume 12 is the hypotenuse

A(squared)+5x5=12x12

A+25=144

144-25=119

Square root of 119=10.9

So I'm going to go with answer 10 in or 11 if you round.

6 0

3 years ago

Maths trig functions <br> please help

Nikitich [7]

1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$ .

We can confirm this algebraically.

$\sin(x) + 2 = -\cos(x) + 3$

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$ . For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$ .