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WARRIOR [948]
3 years ago
9

Look at the diagram. Which of the following is another name for < 1

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0
Another name for ∠<span>1 is
</span><span>A. </span>∠<span>CBA </span>
Mariana [72]3 years ago
4 0

Answer: A. ∠CBA

Step-by-step explanation:

We need three points to write a name of an angle. One point on each ray with the vertex lies in the middle.

There are total three ways to write a name of an angle.

For example :- M and N are points on each ray of an angle with vertex O.

Then, the name of angle can be ∠O, ∠MON or ∠NOM.

From the given picture , it can be observe that A and C are points on each ray of ∠1 with vertex B.

Then, the possible name of ∠1 will be ∠B , ∠ABC or ∠CBA.

Since, ∠CBA in given in option A.

Thus OPTION A is the correct option.

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Como resolver y=1-2.5
oksian1 [2.3K]

I don't really speak Spanish but I'll guess that says

 How do I solve  y = 1 - 2.5

Unless I'm missing something, there's not even any algebra here.  

The way we subtract a bigger number from a smaller number is first we subtract the smaller number from the bigger number the regular way, then we prepend a minus sign.


y = 1 - 2.5 = -(2.5 - 1) = -1.5



7 0
3 years ago
A cylinder has a radius of 12.2 inches and a height of 6.1 inches.
Anna [14]

<span>A cylinder has a radius of 12.2 inches and a height of 6.1 inches. What is the volume of the cylinder to the nearest tenth in³? 

</span><span>C. 2852.3 in³</span>
8 0
3 years ago
Read 2 more answers
50 POINTS PLEASE HELP AND FAST I HAVE BEEN STUCK FOR A WHILE. PLEASE PROVIDE A DECENT EXPLANATION
gayaneshka [121]

Answer:

pearls profs

Step-by-step explanation:

becuse i got the same qwestion and got it wrtei

7 0
3 years ago
I really really really need help!!!!
Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
8 0
3 years ago
Whats 2-2? I'm 8<br> Please help me
jolli1 [7]

Answer:

0

Step-by-step explanation:

2-2=0

6 0
2 years ago
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