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Gwar [14]
3 years ago
12

Classify each angle Name Each Angle in Four Ways

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
7 0
Well the first one is obtuse and more than 90 degrees, the 2nd one is also obtuse and more than 90 degrees, the 3rd one is acute and less than 90 degrees, the 4th one is obtuse and 180 degrees, and the 5th one is a right angle and is 90 degrees
8_murik_8 [283]3 years ago
5 0

Answer:

1) Obtuse angle

2) Obtuse angle

3) Acute angle

You can do 4 and 5. I will give you a hint that acute angles are less than 90 degrees and obtuse angles are more than 90 degrees. And if it is a straight line it is 180 degrees.

6) Angle KLM, MLK, Obtuse angle, and angle 5.

7) Angle DCB, BCD, Obtuse angle, and angle 5.

You can do 8. I will give you a hint that the letters will start from end-middle-end and it can be opposite. If getting confused then look what I did. On muber 6, my I typed angle KLM and angle MLK. Both of them are same but look on how I typed. Also while naming angles, they will be in capital.

Hope this helps, thank you !!

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Alexis buys groceries using her credit card her card has an APR of 19.99% and she must pay at least 5% of the balance at the end
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Answer:

The answer is that she would pay $65.56 in finance charges at the end of the month.

Step-by-step explanation:

Given: APR = 19.99%

Carry Over Balance: $398.97

The APR or Annual Percentage Rate, is calculated daily. You will need to get the daily periodic rate, or DPR, so divide the APR by 365:

19.99% = .1999

.1999 / 365 = .005477 (This is the Approximate DPR, rounded up to .005477)

To get the finance charge, multiply the average daily balance by the DPR and then by 30 days:

398.97 * .005477 * 30 = $65.56 finance charge for this carry over balance, at the end of the month. This assumes that the balance is the average daily balance.

Hope this helps!!  Have a great day!

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3 years ago
Please help me I will crown
Naddika [18.5K]

Answer: c

Step-by-step explanation:

3 0
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which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the l
victus00 [196]
One equation for this would be

y = \frac{41}{16} x-\frac{55}{8}

We start by finding the slope between the two points:

m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}&#10;\\&#10;\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})&#10;\\&#10;\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})&#10;\\&#10;\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

y-y_1=m(x-x_1)&#10;\\&#10;\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})&#10;\\&#10;\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}&#10;\\&#10;\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}&#10;\\&#10;\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{550}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{55}{8}
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