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Rudik [331]
3 years ago
13

2. You have just signed an annual contract with TELUS for a cell phone. The base rate is $22 per month and $0.44 per minute. The

re are no free minutes included with this contract.
a) Write a linear equation in which y represents the total cost for the cell phone per month and x represents the number of minutes spend on the phone each month.
Mathematics
1 answer:
Drupady [299]3 years ago
6 0
Y=0.44x+22 would work I'm pretty sure
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The data is the number of text messages you send a day. find the mean, median, and mode of the data.
Novay_Z [31]

Answer:

6, 8, 10, 12, 18, 20

Mean: 6 + 8 + 10 + 12 + 18 + 20 = 74/6 = <em>12.3</em>

Median: <em>11</em>

Mode: <em>there are no mode in the following listed numbers or data set.</em>

<em />

~hope this helps~

would highly appreciate it if you would mark me the brainliest

4 0
3 years ago
3. Simplify: 72 +8 [180 + 4 {10 + (15 - 45 - 9 x 2)}] simplify ​
garri49 [273]

Answer:

296

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Help please asapppppppp
Schach [20]

Answer:

see below

Step-by-step explanation:

Order of operations is parenthesis, exponents, multiplication, division, addition, subtraction.

since there are no exponents in this problem the next step is -2(2.2)

Simplify the previous step and you get -4.4

final answer is -7.6 - 4.4 = -12

5 0
4 years ago
Solve for z. z/12=6 3/4
Tatiana [17]
6 3/4=6*4+3/4=24+3/4=27/4

z/12=27/4
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3 0
3 years ago
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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
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