Suppose that you have a square pyramid like the one pictured. Which plane section will produce a triangle?

In a package of candy there is a total of 24 yellow pieces. For every 5 green pieces in the package, there are 8 yellow pieces.

IgorLugansk [536]

Answer:

15 green pieces.

Step-by-step explanation:

8x = 24

x = 3.

So you need 3 times the amount of green pieces for every yellow piece.

3 * 5 green pieces = 15 green pieces

7 0

3 years ago

Which is greater,,, 3/6 or 2/5

Karolina [17]

3/6 is the answer

3/6is 0.50 and 2/5 is 0.40

Just divide them

6 0

3 years ago

Read 2 more answers

1. Write the first term of the arithmetic sequence whose first term is 7, and whose common difference is 9.

Elis [28]

Answer:

first term: 7

find the 17th term : 102

fifteent term: 10

Step-by-step explanation:

hababa

6 0

3 years ago

Suppose you have a set of data points with a correlation of 0.85. What is the variance? A. Variance = √ r = 0.92195 B. Variance

TiliK225 [7]

Answer:

B. Variance = r2 = 0.7225

Step-by-step explanation:

Given the correlation coefficient, variance is obtained by squaring the correlation coefficient to obtain what is also know as coefficient of determination. Gives information on the predictive power of the model

8 0

3 years ago

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

2 years ago