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3 years ago

25% of 395 is what?

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

Answer:

98.75

Step-by-step explanation:

25 x 395 = 9875

9875 / 100 = 98.75

hope this helps

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Which represents a quadratic function?

Lera25 [3.4K]

Answer:

Step-by-step explanation:

Okay! So before we do this, first we have to understand the function. The function is,

y = ax^2 + bx + c, where a does not equal to 0.

The answer is not A, since bx is bx^2 which is incorrect.

The answer isn't C, since there isn't an ax^2 or a bc.

The answer isn't D since there a can't be 0.

The only correct answer is B, for it has all the parts.

x^2 + 2x -5.

7 0

3 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:

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

8 0

3 years ago

What is the result of converting 8 gallons to quarts? Rememer, there are 4 quarts in a gallon.

Darya [45]

4 quarts = 1 gallon

32 quarts = 8 gallons

All you have to do is 8 multiplied by 4.

8 * 4 = 32

Hope this helps you! Happy Thanksgiving, here's a turkey!

5 0

3 years ago

Read 2 more answers

Please help its due today! Please show work

Step2247 [10]

Answer:

Step-by-step explanation:

I didn’t know

3 0

3 years ago

What is the solution to this system of equations? {5x+4y=22 3x+4y=10

zlopas [31]

Multiply 2nd equqation by -1 and add to first

5x+4y=22
-3x-4y=-10 +
2x+0y=12

2x=12
divide by 2
x=6

sub back

3x+4y=10
3(6)+4y=10
18+4y=10
minus 18 both sides
4y=-8
divide by 4
y=-2

(x,y)
(6,-2) is solution

6 0

3 years ago

Read 2 more answers