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2 years ago

COMPUTERS The byte is the fundamental unit of computer processing.

Mathematics

1 answer:

Vitek1552 [10]2 years ago

5 0

Answer:

1000x bigger

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PR is represented by which sketch?

solniwko [45]

Answer:

Step-by-step explanation:

The direction of the arrow has to go the same way as the one in the question, from P to R. Also, the question only has one arrow, so the answer only has one arrow.

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3 years ago

It costs$15 to send 3 packages through a certain shipping company Consider the number of packages per dollar.

VMariaS [17]

.2 packages per number bebe :)

4 0

3 years ago

Find the are of a parallelogram that has a base of 7 1/2 and a height of 5 2/7

Amiraneli [1.4K]

Use the formula for the area of a parallelogram:

$\sf A=bh$

Plug in what we know:

$\sf A=(7\dfrac{1}{2})(5\dfrac{2}{7})$

Convert both of them into improper fractions. We do this by multiplying the denominator to the whole number, adding it to the numerator, which becomes our new numerator, and we keep the denominator the same:

$\sf A=(\dfrac{15}{2})(\dfrac{37}{7})$

Now multiply the numerators and denominators together:

$\sf A=\dfrac{15}{2}\cdot\dfrac{37}{7}\rightarrow\dfrac{15\cdot 37}{2\cdot 7}\rightarrow\dfrac{555}{14}$

Convert it back into a mixed number. 14 goes into 555 thirty-nine times. 14 * 39 = 546. 555 - 546 = 9. So we have 39 wholes and 9 left over, or:

$\sf 39\dfrac{9}{14}$

So this is the area of the parallelogram.

6 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

PLS HELP IF YOU CAN!THIS IS DUE IN 30 MINUTES :(

ExtremeBDS [4]

Answer:

Length: 20 or 40 feet

Width: 40 or 20 feet

Step-by-step explanation:

Area = 800 = length*width = x*(60-x)= 60x - x^2

- x^2 + 60x - 800 = 0

x = 20

Or x = 40

So length can be either 20 feet or 40 feet and width can be 40 (60-20) or 20 (60-40)

6 0

3 years ago