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Cloud [144]
3 years ago
7

I need to find the arc! Help

Mathematics
1 answer:
cricket20 [7]3 years ago
3 0
I would say it is 32 on arc CD
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Abbie recorded the temperatures at which two oils freeze: Oil Temperature Cottonseed oil −48.1 °C Coconut oil 24.8 °C Which stat
alekssr [168]

Answer:

The correct answer is option (C).

Coconut oil freezes at a higher temperature than cottonseed oil because 24.8 > -48.1

Step-by-step explanation:

We have been given the freezing temperatures of cottonseed oil and coconut oil. The freezing temperature of cottonseed oil is -48.1 degree celsius and the same for coconut oil is 24.8 degree celsius.

We are supposed to compare the two freezing temperatures. We know that 24.8 is higher than -48.1, therefore, we can conclude that freezing temperature of coconut oil is higher than the freezing temperature of cottonseed oil because 24.8 is more than -48.1. Thus, correct answer is option (C).

4 0
4 years ago
What do you know about these two lines?
ki77a [65]
<h3>Answer: Choice C</h3>

Explanation:

Refer to the y = mx+b form

  • m = slope
  • b = y intercept

The slopes are the only thing we focus on. For the equations y = 6x-1 and y = -6x-4, the slopes are 6 and -6 in that order.

The slopes are not equal, so the lines aren't parallel.

The slopes aren't opposite reciprocals of one another, so they aren't perpendicular either. An example of an opposite reciprocal pairing would be 6 and -1/6. Any pair of nonzero opposite reciprocal slopes always multiply to -1.

Therefore, the two lines are neither parallel nor perpendicular, they just intersect.

6 0
2 years ago
If 9 : x = x:4, then x =
miskamm [114]

Answer:

x = ±6

Step-by-step explanation:

9 : x = x : 4

Cross multiply

36 = x²

Take the square root of both sidesw

x = ±6

5 0
3 years ago
Please solve this question.​
Anon25 [30]

\left(\dfrac{1}{1+2i}+\dfrac{3}{1-i}\right)\left(\dfrac{3-2i}{1+3i}\right)=\\\\\left(\dfrac{1-2i}{(1+2i)(1-2i)}+\dfrac{3(1+i)}{(1-i)(1+i)}\right)\left(\dfrac{(3-2i)(1-3i)}{(1+3i)(1-3i)}\right)=\\\\\left(\dfrac{1-2i}{1+4}+\dfrac{3+3i}{1+1}\right)\left(\dfrac{3-9i-2i-6}{1+9}\right)=\\\\\left(\dfrac{1-2i}{5}+\dfrac{3+3i}{2}\right)\left(\dfrac{-3-11i}{10}\right)=\\\\\left(\dfrac{2(1-2i)}{10}+\dfrac{5(3+3i)}{10}\right)\left(\dfrac{-3-11i}{10}\right)=

\dfrac{2-4i+15+15i}{10}\cdot\dfrac{-3-11i}{10}=\\\\\dfrac{17+11i}{10}\cdot\dfrac{-3-11i}{10}=\\\\\dfrac{-51-187i-33i+121}{100}=\\\\\dfrac{70-220i}{100}=\\\\\dfrac{70}{100}-\dfrac{220i}{100}=\\\\\boxed{\dfrac{7}{10}-\dfrac{11}{5}i}

8 0
3 years ago
HELP PLS EASY POINTS (probably)
DIA [1.3K]

Answer:

C.

Step-by-step explanation:

3 0
3 years ago
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