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2 years ago

Evaluate A2 for A = -3.

Mathematics

2 answers:

4vir4ik [10]2 years ago

6 0

If you are solving A^2 then the answer is 9.
Or If you are solving A•2 then the answer is -6.

Marysya12 [62]2 years ago

6 0

Answer: The required value of A² is 9.

Step-by-step explanation: We are given the following value of a variable A :

$A=-3.$

We are to find the value of $A^2.$

To evaluate $A^2,$ we need to multiply A and A.

Therefore, the evaluation is as follows :

$A^2=A\times A=(-3)\times(-3)=9.$

Thus, the required value of A² is 9.

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ivolga24 [154]

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62 shows up one more time than all the other numbers, and so 62 is your answer.

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2 years ago

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A biologist is researching a newly-discovered species of bacteria. He puts 1,000 bacteria into what he has determined to be a fa

Masteriza [31]

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2 years ago

So I’m not smart so like help meh pleasee<br><br> What is 49% of 200?

____ [38]

Answer:98

Step-by-step explanation:

We assume, that the number 200 is 100%.if 200 is 100%, so we can write it down as 200=100%.We also know, that x is 49% of the output value, so we can write it down as x=49%.Now we have two simple equations:

1) 200=100%

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3 0

2 years ago

Please Help? It is due in 5 mins anyone please?

antiseptic1488 [7]

360°/60 = 6

25 x 6 = 150°

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7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

2 years ago