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exis [7]
3 years ago
5

Determine whether the equation is always sometimes or never true 2(5x+4)=10x+6

Mathematics
1 answer:
Usimov [2.4K]3 years ago
3 0
This equation is NEVER true. Let's solve the equations algebraically by applying the Distributive Property of Multiplication and transposing the terms across the equality sign. The solution is as follows:

<span>2(5x+4)=10x+6
10x + 8 = 10x + 6
10x - 10x = 6 - 8
0 = -2

Therefore, it is not true.</span>
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Step-by-step explanation:

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Multiply, (2x^4-3y^2)(x^2+5y^4)
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Read 2 more answers
Given the weekly demand curve of a local wine producer is p= 50-0.1q, and that the total cost function is c= 1500+ 10q, where q
Mamont248 [21]

Answer:

Step-by-step explanation:

From the given information:

a) To express the weekly profit as a function of price

Cost =C(q) = 1500 + 10q

Revenue = p×q = (50 − 0.1q)×q = 50q - 0.1q²

Revenue = 50q - 0.1q²

Weekly profit = Revenue - Cost

P(q) = (50q -0.1q²) - (1500 + 10q)

P(q)= -0.1 q² + 40 q - 1500  

However, q = 500 - 10 p using p = 50 − 0.1q

P= -0.1 (500 - 10 p)² + 40 (500 - 10 p) - 1500

P= -10 p² + 600 p - 6500

b)

The price at which the bottle of the wine must be sold to realise a maximum profit can be determined by finding the derivative and then set it to 0  

P' = 0

= -20p+600 = 0

20p = 600

p = 600/20

p = $30

c)

The maximum profit that can be made by the producer is:

P= -10(30)² + 600(30) - 6500

P = - 9000 + 18000 - 6500

P = $2500

5 0
3 years ago
784x44=? Step by step.
Triss [41]

Answer:

34496

Step-by-step explanation:

i dont know how 2 explain this but you have to do the traditional way of multiplying PLS MARK BRAINLIEST

7 0
2 years ago
CAN SONEONE PLEASE HELP ME WITH MY MATH PLEASEEEE!!!!!​
Marina86 [1]

Answer:

√190

Step-by-step explanation:

In the figure , there are 2 right angled triangles with a common perpendicular & both the triangles combine to form a new right angled triangle.

Let the triangle with 9 as base be T¹ & Let the triangle with base 10 be T². Let the triangle formed by T¹ & T² be T³.

In T² ,

Hypotenuse = y

Base = 10

According to Pythagorean Theorem ,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Hence, (Perpendicular)² = y^2 - 10^2 = y^2 - 100

In T¹ ,

Perpendicular = \sqrt{y^2 - 100}   (∵ Both T¹ & T² have common perpendicular)

⇒(Perpendicular)² = y^2 - 100

Base = 9

⇒ (Base)² = 9²

Hypotenuse =

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ (Hypotenuse)² = y^2 - 100 + 9^2 .............................................eqn.2

Now in T³ ,

Base = y

⇒ (Base)² = y²

Perpendicular = \sqrt{(y^2 - 100) + 9^2} (∵Perpendicular of T³ = Hypotenuse of T²)

⇒ (Perpendicular)² = (\sqrt{(y^2 - 100) + 9^2})^2= (y^2 - 100) + 81 = y^2 - 19

Hypotenuse = 9 + 10 = 19

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

=> 19^2 = y^2 - 19 + y^2\\\\=> 2y^2 = 19^2 + 19 = 19(19 + 1) = 19*20\\\\=> y^2 = \frac{19*20}{2} = 19*10 = 190\\ \\=> y =\sqrt{190}

3 0
3 years ago
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