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2 years ago

How to do multiplication large numbers.

Mathematics

1 answer:

Keith_Richards [23]2 years ago

7 0

Answer:

The main reason to know the multiplication table is so you can more easily multiply larger numbers. For example, suppose you want to multiply 53 x 7. Start by stacking these numbers on top of another, aligning the ones place. Draw a line underneath, and then multiply 3 by 7. Because 3 x 7 = 21, write down the ones digit (1) and carry the tens digit (2) to the tens column:

Next, multiply 5 by 7. This time, 5 x 7 = 35. But you also need to add the 2 that you carried over, which makes the result 37. Because 5 and 7 are the last numbers to multiply, you don’t have to carry, so write down the 37 — you find that 53 x 7 = 371:

When multiplying larger numbers, the idea is similar. For example, suppose you want to multiply 53 by 47. Be sure to align the stacked numbers by the ones place. (The first few steps — multiplying by the 7 in 47 — are the same, so pick up the next step.) Now you’re ready to multiply by the 4 in 47. But remember that this 4 is in the tens column, so it really means 40. So to begin, put a 0 directly under the 1 in 371:

This 0 acts as a placeholder so that this row is aligned properly.

When multiplying by larger numbers with two digits or more, use one placeholding zero when multiplying by the tens digit, two placeholding zeros when multiplying by the hundreds digit

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lesantik [10]

F(x) is a quadratic. The y intercept, therefore, is equal to the c value.
The y intercept here is -4.
For g(x), you can tell that the y intercept is 0 because that's the value of y when the x value is 0.
For h(x), the chart specifies that when x=0, y=-2, so the y intercept is -2.
Of these three values, 0 is the largest.
Final answer: g(x)

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3 years ago

Kim eats 1/3 of an apple in 1/6 minute.How many minutes will it take

klio [65]

Answer:

0.5 or 3/6 min

Step-by-step explanation:

1/3= 1/6min or 10 sec

to finish the whole apple 3/3 is required

1 part is 10 sec, 3 parts will be 30 secs which is 0.5 /3/6 min.

hope this helps

3 0

3 years ago

4. What is the solution of the system of equations?<br> y=-2x+5<br> y=-2x+20

Norma-Jean [14]

The system of equation, y = -2x + 5 and y = -2x + 20, have no solution.

<h3>How to Find the Solution of a System of Equations?</h3>

The solution to a system of equations is the ordered pair of x and y that makes both equations true.

Given the system of equations,

y = -2x + 5 --> equation 1

y = -2x + 20 --> equation 2

Subtract to eliminate variable "y"

0 = 0 + 25

0 = 25 [this is not true]

Therefore, no solution exist for the system of equations.

Thus, we can conclude that, the system of equation, y = -2x + 5 and y = -2x + 20, have no solution.

Learn more about system of equations on:

brainly.com/question/14323743

#SPJ1

3 0

2 years ago

Please help with this questions. I’ve waisted over 100 points trying to get answers for one assignment just because the only peo

Scrat [10]

Y
-7
-4
-1
2
5

(X,Y)
(-2,-7)
(-1,-4)
(0,-1)
(1,2)
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5 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