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4 years ago

How do you solve this ???

Mathematics

1 answer:

galina1969 [7]4 years ago

3 0

First you have to figure out how much a video game costs and how much a used one costs
Then you plug in the costs of the video games into Janets equation 120=3x+y
(x= video games and y=used video games)
Then you subtract the cost of video games from 120 and then divide that answer by the cost of used video games and that should give you how many used video games she can get

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What is 10 time 100000000000

katrin2010 [14]

1,000,000,000,000 is the answer

7 0

3 years ago

Gina works at a diner. She earns $6 each hour plus tips. I. One week she worked 37 hours a week and earned $43 in tips. How much

FromTheMoon [43]

Answer:

$265

Step-by-step explanation:

The answer is $265 (how much did she make altogether)

The equation is x=6h+6

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4 years ago

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Answer: "B"

Step-by-step explanation:

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4 years ago

The following masses are given in kilograms. Use metric prefixes on the gram to rewrite them so the numerical value is bigger th

vovangra [49]

Answer:

a) 38 mg

b) 230 Exa-g

c) 24 ng

d) 8 Eg

e) 4.2 g

Step-by-step explanation:

Given:

a) 3.8×10⁻⁵ kg

1 kg = 10⁶ mg

Multiply the above value by 10⁶

we get

3.8×10⁻⁵ × 10⁶ = 38 mg

b) 2.3 × 10¹⁷ kg

2.3 × 10²⁰ g

1 exa = 10¹⁸

multiply by 10⁻¹⁸

we get

230 Exa-g

c) 2.4 × 10⁻¹¹ kg

24 × 10⁻⁹ g

1 g = 10⁹ ng

thus,

24 × 10⁻⁹ g × 10⁹ = 24 ng

d) 8×10¹⁵ kg

8 × 10¹⁸ g

1 exa gram = 10¹⁸ g

( 8 × 10¹⁸ ) × 10⁻¹⁸ = 8 Eg

e) 4.2 × 10⁻³ kg

4.2 × 10⁻³ × 10³ = 4.2 g

5 0

3 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

3 years ago