Someone plz help me :(

Mathematics

1 answer:

babymother [125]3 years ago

7 0

Answer:

First one is True, Second is false

Step-by-step explanation:

On Monday the absolute value was 2, on Tuesday the absolute value was 4. 2<4. Hope this helps

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Write the equation of the parabola below

lorasvet [3.4K]

Answer:
Equation is: y = 0.5x² + 0.5x - 3

Explanation:
general form of the parabola is:
y = ax² + bx + c

Now, we will need to solve for a, b and c.
To do this, we will simply get points from the graph, substitute in the general equation and solve for the missing coefficients.

First point that we will use is (0,-3).
y = y = ax² + bx + c
-3 = a(0)² + b(0) + c
c = -3

The equation now becomes:
y = ax² + bx - 3

The second point that we will use is (2,0):
y = ax² + bx - 3
0 = a(2)² + b(2) - 3
0 = 4a + 2b -3
4a + 2b = 3
This means that:
2b = 3 - 4a
b = 1.5 - 2a ...........> I

The third point that we will use is (-3,0):
y = ax² + bx - 3
0 = a(-3)² + b(-3) - 3
0 = 9a - 3b - 3
9a - 3b = 3 ...........> II

Substitute with I in II and solve for a as follows:
9a - 3b = 3
9a - 3(1.5 - 2a) = 3
9a - 4.5 + 6a = 3
15a = 7.5
a = 7.5 / 15
a = 0.5

Substitute with the value of a in equation I to get b as follows:
b = 1.5 - 2a
b = 1.5 - 2(0.5)
b = 0.5

Substitute with a and b in the equation as follows:
y = 0.5x² + 0.5x - 3

Hope this helps :)

7 0

4 years ago

Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .