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1 year ago

Which point on the number line represents Negative three-fourths?

Mathematics

2 answers:

quester [9]1 year ago

8 0

I am not sure but I think The answer is B

Gemiola [76]1 year ago

7 0

Answer:

Step-by-step explanation:

It is 1/4 more than -1, so it is -3/4

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Find the perimeter of the following polygon. Be sure to include the correct unit in your answer.

slava [35]

Answer:

Step-by-step explanation:

All you got to do is add all of them together

8 0

2 years ago

Aidan has 6 (different) bulls and 10 (different) horses living at his animal sanctuary. He needs to place them in a line of 16 p

suter [353]

Answer: 5225472000

Step-by-step explanation:

Given : The number of bulls = 6

The number of horses = 10

Since Aidan needs to place them in a line of 16 paddocks, and the bulls cannot be placed in adjacent paddocks .

Also there are two ways to arrange the group pf bulls and horses.

Then , the number of ways Aiden can place the bulls and horses in the paddocks so that no two bulls are paddocks will be :_

$2\times6!\times10!=5225472000$

Hence, the number of ways Aiden can place the bulls and horses in the paddocks so that no two bulls are paddocks =5225472000

7 0

2 years ago

Rewrite the expression by favoring out the greatest common factor: 9b+12

malfutka [58]

Answer:

3(3b + 4)

the GCF is 3

Step-by-step explanation:

7 0

2 years ago

A merry-go-round can hold 36 people on each 5-minute ride. How many people would have received a ride on the merry-go-round in a

BigorU [14]

Answer:

432

Step-by-step explanation:

The hour have 12 5 min (12*5=60)

So 36*12=432

3 0

2 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

1 year ago