I reallyyyyyyyyyyy need help!

1 answer:

6 0

It lies in quadrant III. If we subtract 360 from 570, then we know that 570° is the same as 210°. Remember that quadrant I has 0-90, II had 90-180, III had 180-270, and IV has 270-360. Then, 570 lies in quadrant III because the corresponding angle, 210, lies in quadrant III between 180 and 270. Hope this helps. :)

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I need help with number 30

yanalaym [24]

Answer:

220

Step-by-step explanation:

5 0

4 years ago

Find the square. (1/4A + 1/4B)^2<br><br> a) 1/16A^2 + 1/8AB + 1/16B^2<br> b) 1/4A^2 + 1/8AB + 1/4B^2

alisha [4.7K]

$(a + b)^2 = a^2 + 2ab + b^2$

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2$

Answer: A.

Another way to solve by factoring 1/4.

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= [\dfrac{1}{4}(A + B)]^2$

$= (\dfrac{1}{4})^2(A + B)^2$

$= \dfrac{1}{16}(A^2 + 2AB + B^2)$

$= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2$

6 0

4 years ago

Read 2 more answers

Last month Jim drove his car 2,718.3 miles. That brought the car’s total mileage to 87,416. What was the car’s mileage before la

stich3 [128]

To solve this problem you must apply the proccedure shown below:

1. You have that Jim drove the car 2,718.3 miles for a total mileage of 87,416.

2. Then, to calculate the mileage before last month, you only need to substract the total mileage given in the exercise above and the mileage drove last month, as following:

$87,416miles-2,718.3miles=84,697.7miles$