Find the limit of the function algebraically. limit as x approaches negative six of quantity x squared minus thirty six divided by quantity x plus six.

Answers

Answer 1

Answer: $lim_(x \rightarrow -6) (x^2-36)/(x+6) \n =lim_(x \rightarrow -6) ((x-6)(x+6))/(x+6) \n =lim_(x \rightarrow -6) (x-6) \n =-6-6=-12$

Answer 2

Answer:

Answer:

-12

Step-by-step explanation:

$\lim_(x \to \ -6) (x^2-36)/(x+6)$

We factor the numerator and try to simplify the fraction as much as we can.

$x^2-36 = x^2 - 6^2$

Apply a^2 - b^2 formula (a+b)(a-b)

$x^2-36 = x^2 - 6^2=(x+6)(x-6)$

$\lim_(x \to \ -6) ((x+6)(x-6)/(x+6)$

Cancel out x+6 from top and bottom

$\lim_(x \to \ -6)(x-6)$

Plug in -6 for x

-6-6 = -12

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Write the ratio in lowest terms. 20:14a. (7/10)
b. (20/14)
c. (14/20)
d. (10/7)

Answers

when putting ratios, fractions, expressions, etc into lowest terms you take out the gcf.

in this case the gcf of both 20 and 14 is 2.

20/2:14/2 = 10:7

the answer is d

to get a ratio in lowest terms, you have to divide both sides by the same number. what number fits into both? 2 does. so 20/2=10, and 14/2=7. this means that (d) is correct.
does that help?

Given the sequence 8, 16, 32, 64, ..., which expression would give the thirteenth term?

Answers

8, 16, 32, 64,…
×2 ×2 ×2

$a_(n) = a_(1) * r^(n - 1)$
$a_(n) = 8 * 2^(13 - 1)$
$a_(n) = 8 * 2^(12)$
$a_(n) = 8 * 4096$
$a_(n) = 32768$

   First Term: 8
      Second Term: 16
      Third Term: 32
      Fourth Term: 64
       Fifth Term: 128
      Sixth Term: 256
   Seventh Term: 512
   Eighth Term: 1024
   Ninth Term: 2048
      Tenth Term: 4096
   Eleventh Term: 8192
   Twelfth Term: 16384
Thirteenth Term: 32768

$The\ expression\ that\ would\ give\ the\ thirteenth\ term\ of\ the\ problem\ would\ be\ the\ equation\ of\ the\ geometric\ sequence,\ which\ is\ equal\ to\ a_(n) = a_(1) * r^(n - 1).$

Please help me this assignment is due todayyy it is 12 points

Answers

Answer:

Step-by-step explanation:

the range is the y-values

The range is the Y values, so answer B.

What can I multiple -8x-4y-3z=11 so that the X variable Can eliminate with 6x+8y-7z=-63

Answers

Answer:

fhk frrhjklnby

Step-by-step explanation:

g pay no of the month of November month of the year end of the day is a very happy birthday to you and your family members of the world cup final in the world cup today and today is the last date of the day of the day and the day I was in the same time for you and your family members and all the best for your future

Ten to the fourth power as a product of tens

Answers

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Need help really bad please

Answers

Problem 1

Answer: Choice B) angle1 = 26, angle2 = 64

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Explanation:

Complementary angles add to 90

(angle1)+(angle2) = 90

(5x+6)+(68-x) = 90

4x+74 = 90

4x = 90-74

4x = 16

x = 16/4

x = 4

So,

angle1 = 5x+6 = 5(4)+6 = 26

angle2 = 68-x = 68-4 = 64

Note how

angle1+angle2 = 26+64 = 90

to help check our answer

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

Problem 2

Answer: Choice D) BC, AC, AB

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Explanation:

The shortest side of a triangle is always opposite the smallest angle.

The longest side of a triangle is always opposite the largest angle.

Based on those two facts, we can say the shortest side is BC (since angle A = 53 is the smallest angle) and the longest side is AB (since angle C = 70 is the largest)

The order from smallest to largest is

BC, AC, AB

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

Problem 3

Answer: Choice D) B < C < A

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Explanation:

Use the reverse of the idea from problem 2. We know that AC is the smallest side which must mean that angle B is the smallest angle. Since BC is the largest side, angle A must be the largest angle.

The order of the angles is B < C < A

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

Problem 4

Answer: Choice A) angle 5 = 50, angle 8 = 130

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Explanation:

The same side exterior angles must be supplementary to have parallel lines.

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

Problem 5

Answer: Choice A) angle G is the smallest

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Explanation:

Use the same idea as problem 2. The smallest angle is opposite the smallest side. The smallest side is 48, so angle G being opposite of that side, is the smallest angle.

Angle O would be the largest as it is opposite the largest side DG = 70

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