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4 years ago

Solve the two-step equation. − 2 3 x – 21 4 = 27 4

Mathematics

2 answers:

svetoff [14.1K]4 years ago

7 0

Answer:

Step-by-step explanation:

I'm going to assume that the problem is -23x -214 = 274

And in that case...

1. First isolate the x-value by adding 214 to both sides of the equation:

-23x = 488

2. Then, divide both sides by -23 to truly get x alone:

x = $-\frac{488}{23}$ or -21.22

defon4 years ago

7 0

Answer:

x = -18

Add the opposite of the constant on the left:

-2/3x = 21/4 +27/4 = 48/4 = 12

Multiply the equation by the inverse of the coefficient of x:

x = (-3/2)(12) = -18

The solution is x = -18.

Step-by-step explanation:

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Rewrite the fraction as a decimal -19/50

Mama L [17]

Rewrite the fraction as a decimal
-19/50 = -0.38

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3 years ago

The price of a cd was decreased by 20% to £7.68. What was the price before the decrease?

aleksandrvk [35]

Answer:

£9.6

Step-by-step explanation:

x = the original price of a CD

£x = 100% of the original price

The price of a CD was decreased by 20% to £7.68.

This means:

£7.68 = 100% - 20%

£7.68 = 80% of the original price

From this, we will find 1% of the original price.

£7.68 ÷ 80 = 1%

£0.096 = 1%

Since the original price ( x ) = 100% of the original price, we will find 100% of the original price.

£0.096 × 100 = 100%

£9.6 = 100%

Therefore, the original price of a CD = £9.6

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4 years ago

On the way to the lake you stop at a nearby shop to buy some food and drink. You pick up a 1.5 litre bottle of water. One friend

melomori [17]

Hello,

Sadly, I can't help you, because it seems as the question has been cut off. If you can update your question with the rest of what is asked, or the different choices, I'd be more than happy to assist you in any way I can!

I will update my answer then.

Thank you!

7 0

3 years ago

I don’t u sweat and this

Aloiza [94]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

8 0

4 years ago