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3 years ago

Find the value of x in each case: PU ∩ RT = S, Q ∈ PR

Mathematics

1 answer:

Eddi Din [679]3 years ago

3 0

Answer:

x=34

Step-by-step explanation:

When we look triangle SQR, we have angles 3x and 68, we know that sum angles of triangle is 180.

<R=180-(3x+68)

Now, we look at the PQS:

<PQS=180-68=112°

Then :

<PSQ=180-(112+x)=68-x

Then we have :

Angles PSR= 68-x+3x=68+2x and angles TSU=4x.

We have that

PSR=TSU

68+2x=4x

4x-2x=68

2x=68

x=34

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Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

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$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

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