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Effectus [21]
3 years ago
14

Graph each function y= -x2

Mathematics
1 answer:
alina1380 [7]3 years ago
8 0
Y=-x^2 is just an inverted parabola, opening downward, with a vertex, its absolute maximum at (0,0) and decreasing without bound as x approaches ±oo
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Find the area of this rhombus
Vikentia [17]
<h2>Hello!</h2>

The answer is:

The area of the rhombus is equal to 64 squared inches.

area=64in^{2}

<h2>Why?</h2>

Since we already know half of the length of the diagonals of the rhombus, we can calculate the area of the rhombus using the following formula:

area=\frac{diagonal_{1}*diagonal_{2}}{2}

From the image we can see that:

diagonal_{1}=8in+8in=16in

diagonal_{2}=4in+4in=8in

So, substituting, we have:

area=\frac{16in*8in}{2}=\frac{128in^{2} }{2}=64in^{2}

Hence, we have that the area of the rhombus is equal to 64 square inches.

Have a nice day!

7 0
3 years ago
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1 What is the value of this expression?<br> 2 +4² x 4 - 12
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<u><em>Tell Me If Somethings wrong with my answer! </em></u>

7 0
3 years ago
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The area of a rectangle is x 2 – 3x – 28. If the<br><br>length is x + 4, find the width.
Hatshy [7]

Answer:

-4 square units

Step-by-step explanation:

a = l × b

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x2 - 3x - 28 = x2 + 4x

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Answer:

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Step-by-step explanation:

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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.

3 0
3 years ago
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