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3 years ago

What is seven minus eight over three?

2 answers:

7 0

If u want to know whats (7-8)/3 is you should just say 7-8 = -1 so the answer is -1/3 but if you mean like 7 - ( 8/3 ) so you should say (21-8) /3 = 13/3 :)))
i hope this be helpful
* marry Christmas *

Anna007 [38]3 years ago

5 0

Since 7= 21/3 and you are subtracting 8/3, you subtract 8 from 21. That gives you 13/3 or 4 1/3 depending on how you look at it.

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1 + tanx / 1 + cotx =2

Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer: x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0

3 years ago

I need help with this pleaseee!

ololo11 [35]

Probably B

STEP BY STEP

7 0

3 years ago

What does the A equal to in a + (-7) = 3

Korolek [52]

A = 10

Step by step

3+7=10
10+(-7)=3

7 0

2 years ago

Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> 4x3 + ln y2 + 2y = 2x

lisov135 [29]

Answer:

$\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}$

Step-by-step explanation:

Given function : 4x³ + ln(y²) + 2y = 2x

on differentiating both sides with respect to 'x', we get

$\frac{d(4x^3 + ln(y^2) + 2y)}{dx}= \frac{d(2x)}{dx}$

$12x^2+\frac{2}{y}(\frac{dy}{dx})+2(\frac{dy}{dx})=2$

$12x^2+(\frac{2}{y}+2)\frac{dy}{dx}=2$

$(\frac{2+2y}{y})\frac{dy}{dx}=2-12x^2$

$\frac{dy}{dx}=\frac{y(2-12x^2)}{2+2y}$

$\frac{dy}{dx}=\frac{2y(1-6x^2)}{2(1+y)}$

$\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}$

3 0

3 years ago

Rei is barricading a door to stop a horde of zombies. She stacks boxes of books on a table in front of the door. Each box weighs

Alinara [238K]

Answer: The answer is $\textup{W}=70+30x.$

Step-by-step explanation: Given that Rei is barricading a door to stop a horde of zombies by satcking boxes of books on a table in front of the door.

The weight of each box is 30 kilograms and the weight of the table with 8 boxes on the top is 310 kilograms.

So, if 'g' denotes the weight of the table alone, then we have

$\textup{g}+8\times 30=310\\\\\Rightarrow \textup{g}+240=310\\\\\Rightarrow \textup{g}=310-240\\\\\Rightarrow \textup{g}=70.$

Therefore, the weight of the table without boxes is 70 kilograms.

Now, if 'W' is the total weight of the barricade, then it can be written as a function of 'x', the number of boxes Rei stacks on the table. The expression is as follows -

$\textup{total weight of the barricade}=\textup{weight of the table alone}+30\times \textup{number of boxes}.$

Thus, the required function is

$\textup{W}=70+30x.$

6 0

3 years ago

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