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Anna11 [10]
2 years ago
7

How many subjective functions are there from {1,2,3,4} to {1,2,3}

Mathematics
1 answer:
Ann [662]2 years ago
4 0

Answer:

IM ON THE SAME ONEEE let me know when you have it

Step-by-step explanation:

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Find the equation of the circle that is shifted 5 units to the left and 2 units down from the circle with the equation x^2+y^2=1
marysya [2.9K]
Shifting a circle results to changes in the coordinates of the circle. For instance, if the coordinates of the center of the circle is taken to be (0,0), the new coordinates will be [(0+5),(0+2)] after shifting. The equation of the circle will also change with the same margin.

That is, the new equation will be;
(5+x)^2+(2+y)^2 =19

Notice, only the coordinates changes.
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3 years ago
Thomas has 12 more marbles than twice the number of marbles Andrew has.
Thepotemich [5.8K]

Answer:

A.

x + 2 + 12

Step-by-step explanation:

Thomas (t) = Andrew (x) * 2 + 12

3 0
2 years ago
What is 9%+278.045 or 278.045+9%
rodikova [14]
The answer to this equation would be 278.135
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3 years ago
Which angles are corresponding angles?
konstantin123 [22]

Answer:

A. 7 and 3

&

B. 1 and 5

Step-by-step explanation:

Corresponding angles are a pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal.

The answers are:

•8 and 4

•5 and 1

•7 and 3

•6 and 2

I hope this helps! I'm sorry if it's wrong.

3 0
3 years ago
Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
2 years ago
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