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gladu [14]
2 years ago
15

\frac{3}{2}+\frac{2x+1}{6}=-1 2 3 ​ + 6 2x+1 ​ =−1

Mathematics
2 answers:
malfutka [58]2 years ago
5 0

Answer:

\frac{3}{2}  +  \frac{2x + 1}{6}  =  - 1 \\ 3 +  \frac{2x + 1}{3}  =  - 2 \\  \frac{9 + 2x + 1}{3}  =  - 2 \\ 10 + 2x =  - 6 \\ 2x =  - 16 \\ x =  - 8

Mamont248 [21]2 years ago
4 0
The answer is x = -8
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Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel
nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

\theta=\frac{\pi}{3}

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be=r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j

By using cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Force along the plane will be=\mid F_x\mid=F\cdot r

Force along the plane will be =\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3N

By using i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0

Therefore, force along the plane=\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)

2.The vector perpendicular to the plane=r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j

The force perpendicular to the plane=\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

The force perpendicular to the plane=4N

Therefore, F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=2\sqrt 3i-6j-2\sqrt3 i-2j=-8j

Hence,sum of two component of forces=Total force.

6 0
3 years ago
Total profit P is the difference between total revenue R and total cost C. Given the following​ total-revenue and​ total-cost fu
gavmur [86]

Answer:

The profit function is:

P(x)=-x^2+1280x-3300

The maximum value is 406, 300 occurring when x = 640.

Step-by-step explanation:

The revenue function is:

R(x)=1300x-x^2

And the cost function is:

C(x)=3300+20x

Then the total profit function will be:

P(x)=R(x)-C(x)=(1300x-x^2)-(3300+20x)=-x^2+1280x-3300

This is a quadratic function.

Therefore, the maximum value of the total profit will occur at its vertex point.

The vertex of a quadratic is given by:

\displaystyle \Big(-\frac{b}{2a}, f\Big(-\frac{b}{2a}\Big)\Big)

In this case, a = -1, b = 1280, and c = -3300.

Then the point at which the maximum profit occurs is at:

\displaystyle x=-\frac{1280}{2(-1)}=640

And the maximum profit will be:

P(640)=-(640)^2+1280(640)-3300=406300

5 0
3 years ago
Read 2 more answers
X^4 − 3x^2 − 90 x= -3
Ahat [919]

Answer:

-36

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

x⁴ - 3x² - 90

x = -3

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em>:                                                                                                  (-3)⁴ - 3(-3)² - 90
  2. Exponents:                                                                                                        81 - 3(9) - 90
  3. Multiply:                                                                                                             81 - 27 - 90
  4. Subtract:                                                                                                            54 - 90
  5. Subtract:                                                                                                            -36
8 0
3 years ago
If f(n) = n 2 - n, then f(-4) is _____.
stich3 [128]

Answer:

susgidhkdldhoxhljxjxyjsjtztxyxiuttxixtidyfyxkgk

Step-by-step explanation:

odyxixfjxkcyickcgixgjx

7 0
3 years ago
Read 2 more answers
Please I need help with this!!!!!!!
inysia [295]

Answer:

Step-by-step explanation:

1). Step 4:

   x=5^{\frac{4}{3}}=(5^4)^{\frac{1}{3}}

   x=\sqrt[3]{5^4} [Since, a^{\frac{1}{3}}=\sqrt[3]{a}]

   x=\sqrt[3]{5\times 5\times 5\times 5}

   Step 5:

   x=\sqrt[3]{(5)^3\times 5}

   x=\sqrt[3]{5^3}\times \sqrt[3]{5}

2). He simplified the expression by removing exponents from the given expression.

3). Let the radical equation is,

   (3x-1)^{\frac{1}{5}}=2

   Step 1:

   (3x-1)^{\frac{1}{5}\times \frac{5}{1} }=2^{\frac{5}{1}}

   Step 2:

   (3x-1)=2^5

   Step 3:

   3x=32+1

   Step 4:

   x=11

4). By substituting x=11 in the original equation.

   (3\times 11-1)^{\frac{1}{5}}=(32)^\frac{1}{5}

                         =(2^5)^\frac{1}{5}

                         =2

There is no extraneous solution.

3 0
2 years ago
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