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ch4aika [34]
3 years ago
15

Which expression is equivalent to -5v+(-2)+1+(-2v)

Mathematics
1 answer:
notsponge [240]3 years ago
5 0

Answer:

Step-by-step explanation:

first try to simp.

-5v+(-2)+1+(-2v)

you'd get -7v-1

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The ratio of 3 to 4 can be written in all the following ways except _______. 4/3 3/4 3:4 three-fourths
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Read 2 more answers
X+2&gt;7 write the solution set in interval notation
REY [17]

Answer:

(5, infinitysymbol)

Step-by-step explanation:

First solve the inequality. Subtract 2 from both sides.

x + 2 > 7

x > 5

So that is one way of writing the answer and it is hopefully kind of understandable. X>5 means all the numbers greater (bigger) than 5, forever to infinity.

Interval notation is a way of writing a set or group of numbers. Interval notation uses ( ) parenthesis or [ ] square brackets. Then two numbers go inside with a comma in between. The first number is where the set of numbers start and the second number is where the set ends. You always put parenthesis around the infinity symbol or negative infinity symbol. You only use a square bracket if the inequality symbols have the "or equal to" underline under the > or <.

So x > 5 in interval notation is:

(5, infinitysymbol)

This shows that 5 is not included in the solution; and all the numbers forever bigger than five are solutions as well.

5 0
1 year ago
How are the divisibility rules involved with the simplification of fractions?
Rashid [163]
Because when you simplify a fraction you divide the number on bottom be the top number to simplify the fraction and make it a mixed number
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3 years ago
What number ten more than 985
Ksenya-84 [330]

\text{Hey there!}\\\text{What number ten more than 985?}\\\\\text{In order for you to find the number that's 10 more than 985. you can}\\\text{add 10 plus  the number 985}\\\\\text{So, 985 + 10 = ? }\\\\\text{? = the mystery number}\\\\\text 10 + 985 = 995}\\\\\boxed{\boxed{\text{Answer: 995}}}\checkmark

\text{We added because you put the word MORE in the sentence}\\\text{And more can mean GO UPWARD}

\text{If you were to say less than you would have to subtract. Since,}\\\text{you are going downward }

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6 0
3 years ago
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
3 years ago
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