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Natalija [7]
1 year ago
9

Can anyone identify the center & radius of these

Mathematics
1 answer:
kobusy [5.1K]1 year ago
6 0

Answer:

43 is the answer to your question

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Compare 4 x 105 and 5 x 103.
german
Your answer should be B !
6 0
3 years ago
Adding which of the following ordered pairs to the set [(0, 1), (2, 4), (3,5)) would make it not a function?
Kitty [74]

Answer:

(0,7)

Step-by-step explanation:

For a set of ordered pairs to be a function, there needs to be "ONE" value of y for each unique x value.

Thus, we see that the 3 ordered pairs given have x values of 0, 2, and 3.

Now, we look in the answer choices and find which ones are 0, 2, or 3 in x-coordinate.

That would be only (0,7). Now the original set has x = 0 corresponding to y = 1

Is that true for (0,7)?? NO!

x = 0 will now correspond to "TWO" values of y.

THis will make this NOT A FUNCTION.

So, (0,7) is correct.

3 0
3 years ago
Pretend that you can have 100 of any one thing. What would you choose?why?
Shkiper50 [21]
I would like 100 like 100 things of food because i can feed the food to the homeless
8 0
3 years ago
Read 2 more answers
What is the percentage of 48g of fat out of a 70g allowance
Elenna [48]
<span>48/70=0.6857
</span><span>0.6857x100=68.57%</span>
5 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.
Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

                                     AD+DB=3

                                       x+DB = 3

                                        DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

BC^{2} = DB^{2} +CD^{2}

BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}

BC^{2} =(3-x)^{2} +2

Since given AD=BC,let us plugin BC=x in above step.

x^{2} =(3-x)^{2} +2

x^{2} =9-6x+x^{2} +2

6x=11

x=\frac{11}{6}

Now we know AD=x=\frac{11}{6} and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

AC^{2} =AD^{2} +DC^{2}

AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}

AC^{2} =\frac{193}{36}

AC=\sqrt{\frac{193}{36} } = 2.315cm

3 0
3 years ago
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