All categories

3 years ago

0.35, 0.68, 0.20, 0.31 Order the numbers greatest to least.

Mathematics

2 answers:

Artyom0805 [142]3 years ago

6 0

Answer:0.68,0.35,0.31,0.20

Step-by-step explanation:

katovenus [111]3 years ago

5 0

Answer:

0.68, 0.35, 0.31, 0.20

Step-by-step explanation:

Well I look at it by 68 is bigger than 35 so that's first. Then 35 is second. So 31 is bigger than 20, so 20 would be last.

You might be interested in

Solve the system of linear equation 8X plus 5Y equals 18 and 6X plus Y equals -2 by using the linear Combination method Emmaus d

Hunter-Best [27]

$\text{The solution is } x = \frac{-14}{11} \text{ and } y = \frac{62}{11}$

Solution:

Given system of equations are:

8x + 5y = 18 ------- eqn 1

6x + y = -2 -------- eqn 2

Multiply eqn 2 by -5

-30x -5y = 10 ----- eqn 3

Add eqn 1 and eqn 3 so that y terms gets eliminated

8x + 5y = 18

-30x -5y = 10

( + ) --------------

-22x = 28

Divide both sides by -22

$x = \frac{28}{-22}\\\\x = \frac{-14}{11}$

Substitute the x value in eqn 1

$8(\frac{-14}{11}) + 5y = 18\\\\\frac{-112}{11} + 5y = 18\\\\5y = 18 + \frac{112}{11}\\\\5y = \frac{310}{11}\\\\y = \frac{62}{11}$

$\text{Thus the solution is } x = \frac{-14}{11} \text{ and } y = \frac{62}{11}$

8 0

3 years ago

A model of a house has been drawn on a coordinate grid. One corner of the house has

sp2606 [1]

Answer:

$Point(9,1)$

Step-by-step explanation:

Given

Point: (6,3)

Required

Translate 2 units down and 3 units left

Taking the translation 1 after other

When a function is translated down, only the y axis is affected;

2 units down implies that, 2 be subtracted from the y value.

The function becomes

$Point: (6, 3 - 2)$

$Point: (6, 1)$

3 units right implies that, 3 be added tothe x value.

The function becomes

$Point(3+6,1)$

$Point(9,1)$

Hence;

Option D answers the question

7 0

3 years ago

Write an expression for 20 times the sum of 4 and 2

marta [7]

Answer:

20x6

Step-by-step explanation:

20x (x stands for multiply/times)

6(sum/add 4+2)

20x6

Hope this helps.

8 0

2 years ago

Read 2 more answers

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$