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3 years ago

What is the difference? Express your answer in lowest terms. 3/8 + 5/16 =

Mathematics

2 answers:

tekilochka [14]3 years ago

8 0

6/16+5/16= 11/16 since you need common denominators

Hitman42 [59]3 years ago

4 0

The answer is .6875 i hoped this helped you

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Please help 100 points!!

Naddik [55]

Answer:

Looking at the first question, it's asking what best describes the probability of tossing a number less than 6 on a number cube that has 6 numbers. Impossible means that it will never land on it, for example asking what the probability of landing on 7 is. Unlikely is something that doesn't happen often. The best option that fits our scenario is option C, likely.

Looking at the second question, it's asking what the probability that the teacher chooses a girl in his class. There are 15 girls and a total of 27 students in the class so we take the probability by doing 15/27. We can narrow both the numerator and the denominator using 3 which gives us 5/9. Therefore, the best option that fits our scenario is option C, 5/9.

Finally, looking at the last question, it's asking what the theoretical probability that the coin will land on heads on the next toss. Theoretical probability doesn't consider how much times Murray tossed the coin, the only thing it cares about is what the actual probability of tossing a coin is. Therefore that makes it a 50% chance of landing on a heads and a 50% chance of landing on a tails. The best option that first our scenario is option B, 1/2.

<u><em>Hope this helps! Let me know if you have any questions</em></u>

7 0

2 years ago

Read 2 more answers

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

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

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

3 years ago

Find the absolute extreme of the function on the closed interval. y=9e^xsinx , [0,pi]

Molodets [167]

For convenience sake, I will let $y=f(x)=9e^{x}\sin x$

First, we evaluate the function at the endpoints of the interval.

$f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0$

Then, we need to find the critical points.

We can start by taking the derivative using the power rule.

$f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)$

Setting this equal to 0,

$9e^x (\cos x+\sin x)=0$

Since $9e^x > 0$ , we can divide both sides by $9e^x$ .

$\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}$

$f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}$

So, the absolute minimum is $\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}$ and the absolute minima are $\boxed{(0,0), (\pi, 0)}$