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3 years ago

— 5х – 4у = -8 What’s the answer

Mathematics

2 answers:

Arisa [49]3 years ago

5 0

You first move the constant to the left so it’s
-5x-4y+8=0
then you change the signs
5x+4y-8=0
that’s the answer

Elden [556K]3 years ago

4 0

Answer:

y = -x + 2

Step-by-step explanation:

— 5х – 4у = -8

-4y = 4x - 8

4y = -4x + 8

y = -x + 2

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Answer:

$\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

Step-by-step explanation:

Find the first implicit derivative using implicit differentiation

$2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}$

Use the substitution of dy/dx to find the second derivative (d²y/dx²)

$\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

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2 years ago

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali

Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

$W =$ Wednesday

$T =$ Thursday

$F =$ Friday

$S =$ Saturday

Their corresponding probabilities are

$P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1$

Since $Y$ = number of days beyond Wednesday that it takes for both magazines to arrive(so possible $Y$ values are 0, 1, 2 or 3)

The possible number of outcomes are therefore $4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)$

The values associated for each of the outcomes are as follows:

$Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3$

The probability mass function of $Y$ is,

$P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19$

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