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3 years ago

6 x 20 = 6 x _ tens= _tens = ____

Mathematics

1 answer:

Gekata [30.6K]3 years ago

8 0

6x20=6x2 tens=12tens=120

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What is the volume of the right rectangular prism?

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250 m because you multiple the base by the height

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2 years ago

For exercises 1 and 2, translate each phrase into an algebraic expression.

Tanzania [10]

Hello There!

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4 years ago

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

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3 years ago

What is the y-intercept of the function f(x)=4 – 5x? A:-5 B:-4 C:4 D:-5

Bogdan [553]

Answer: option C is the correct answer

Step-by-step explanation:

The y-intercept of the function is the point on the y or vertical axis at which the straight line cuts through it. The value on the y axis is read off as the y-intercept .

The equation of a straight line is modelled using the slope - intercept equation of

y = mx + c

Where m is the slope( change in value of y/ change in value of x.

c = the y-intercept

Looking at the function given,

f(x)=4 – 5x? Where f(x) is the same as y in our slope intercept equation.

Rearranging the function, it becomes

f(x)= – 5x +4

The intercept is 4

Option C is the correct answer.

7 0

4 years ago

0.73 as a fraction. please

Shalnov [3]

Answer:

73/100

Step-by-step explanation:

.73 is 73 hundredth which is also 73/100.

3 0

3 years ago

Read 2 more answers

6 x 20 = 6 x ___ tens= ___tens = ____

6 x 20 = 6 x _ tens= _tens = ____