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2 years ago

Please help me ! thank you

Mathematics

1 answer:

Tpy6a [65]2 years ago

4 0

1/2^-2 Can be written as 2^2

Therefore , 2×2=4

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PLEASE HELP.

pentagon [3]

Let x represent the number of shirts. Let y represent the number of pens.

If shirts are on sale for $11.99 each, then x shirts cost $11.99x.

If pants are on sale for $12.99 each, then y pants cost $12.99y.

The total cost is $(11.99x+12.99y).

Sarah can spend up to $65. Then an inequality that represents this situation is

11.99x+12.99y≤65 (this inequality holds when Sarah can spend $65 too)

11.99x+12.99y<65 (this inequality holds when Sarah can spend less than $65).

3 0

3 years ago

0.43 as a fraction ?

lukranit [14]

Answer:

43/100

Step-by-step explanation:

6 0

2 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

2 years ago

Six less than 5 times a number is 24,<br> i just need the equation

valentina_108 [34]

Answer:

Step-by-step explanation:

5x-6=24

3 0

3 years ago

How do you solve this using substitution <br> -8x-y=-30<br> Y=12x-50

never [62]

Step-by-step explanation:

substitude the y value into y in the first equation

-8-(12x-50)=30

-8-12x+50=30

you see it?

then solve

-8+50-30=12x

you got the value = 12x

then you divide with 12

finally you get x

7 0

3 years ago