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2 years ago

Write an absolute value inequality for the graph below.use x for your variable.

Mathematics

1 answer:

Alenkasestr [34]2 years ago

7 0

Step-by-step explanation:

The magnitude of x is equal or less than 7.

=> We have |x| <= 7.

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Plan out an experiment that you could use to demonstrate experimental and theroretical probability

Murrr4er [49]

Answer:

Also, the more trials that you conduct in your experiment, the closer your calculations will be for the experimental and theoretical probabilities. The theoretical probability is 8.3% and the experimental probability is 4%. Although the experimental probability is slightly lower, this is not a significant difference.

Step-by-step explanation:

5 0

2 years ago

Which of the following is not a perfect square a:50 b:49 c:121 d:1

Lostsunrise [7]

50 is not a perfect square because it does not have a perfect square root where as 49 is 7, 121 is 11, and 1 is 1

6 0

2 years ago

Read 2 more answers

Which ordered pair is a solution of the equation? y=8x+3y=8x+3

Mrac [35]

Multiply the first equation by 4 (so that both equations will have 8x terms) and then subtract:

8x-20y = -24

8x +3y = 68

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

0 -23y = -92

Now divide both sides by -23:

y = 4

Find x by plugging y=4 into either equation:

8x +3(4) = 68

8x = 68 - 12

8x = 56

x = 7

So the answer is ordered pair is (7,4)

3 0

2 years ago

Read 2 more answers

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

The following is the recipe to make one dozen muffins: 1 cups flour; 2 cups sugar; 2 tsps. Baking powder; 1 egg; 1 cup Crisco; 1

ivolga24 [154]

1.5 cups of flower, 3 cups sugar, 3 tbsp. baking powder, 1.5 eggs, 1.5 cups of Crisco, 1.5 cups of skim milk (hopefully not skin milk), and 1.5 cups of blueberries. Because 18 is 12 x 1.5 you just multiply all the given numbers by 1.5.

4 0

3 years ago

Write an absolute value inequality for the graph below.use x for your variable.​

Write an absolute value inequality for the graph below.use x for your variable.