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2 years ago

Help!!! 1/3x - 3/4 + 5/6x = -2

Mathematics

1 answer:

Alex73 [517]2 years ago

3 0

Answer:

x=-15/14

Step-by-step explanation:

1/3x+5/6x=-2+3/4

7/6x=-5/4

x=-15/14

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Step-by-step explanation:

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1 year ago

What is 457.8 divide 70 use number sense to help you

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There is a 70% chance that your car will get stuck in the snow during the next big snow fall. Given that you are already stuck i

nikdorinn [45]

Answer:

63%

Step-by-step explanation:

This is a problem of conditional probability.

The two events that are given are:

Car stuck in the snow - Let it be event S. P(S) = 70% = 0.70
Require a tow truck - Let it be event T.

We have to find the probability of being stuck in the snow AND requiring a tow truck which can be given as P(S and T)

We are also given the conditional probability, which is P(T | S) = 90% = 0.90

Using the given formula for our case we can modify the formula as:

$P(T|S)=\frac{P(S \cap T)}{P(S)}$

$0.90=\frac{P(S \cap T)}{0.70}\\\\ P(S \cap T)=0.90 \times 0.70\\\\ P(S \cap T)=0.63$

Therefore, there is 63% (0.63) chance that you will get stuck in the snow with your car AND require a tow truck to pull you out

4 0

3 years ago

In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x

Alexus [3.1K]

Answer:

Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

Step-by-step explanation:

Given Differential Equation,

$y"-2y'+y=xe^x$ ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

$\implies\:y(x)=y_c(x)+y_p(x)$

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

$y_c(x)=C_1e^x+c_2xe^x$

Now for particular solution,

let $y_p(x)=Ax^3e^x$

$y'=Ax^3e^x+3Ax^2e^x$

$y"=Ax^3e^x+6Ax^2e^x+6Axe^x$

Now putting these values in (1), we get

$Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x$

$Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x$

$6Axe^x=xe^x$

$6A=1$

$A=\frac{1}{6}$

$\implies\:y_p(x)=\frac{x^3e^x}{6}$

Therefore, Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

8 0

3 years ago