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3 years ago

What is the solution to this linear system?

Mathematics

1 answer:

Drupady [299]3 years ago

3 0

Solution set is (-1,3)

Hence Option D is correct

Step-by-step explanation:

We need to solve linear system to find the solution.

$x+y=2\,\,\,(1) \\3x-2y=-9\,\,\,\,(2)$

Multiply equation (1) with 2 and add both equations.

$2x+2y=4\\3x-2y=-9\\-------\\5x= -5\\x=\frac{-5}{5}\\ x=-1$

Putting value of x in equation(1)

$x+y=2\\-1+y=2\\y=2+1\\y=3$

So, x=-1 and y=3

So, Solution set is (-1,3)

Hence Option D is correct.

Keywords: System of solutions

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Sloan [31]

The degenerate conic that is formed when a double cone is sliced at the ap-ex by a plane parallel to the base of the cone is a Point.

<h3>What degenerate conic is formed?</h3>

When a plane that is parallel to the base of a double cone is used to slice the ap-ex, the conic section formed is a circle.

Circles lead to a Point degenerate conic being formed because a single point will be formed on the double cone that separates the shape.

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2 years ago

Stefan and his friends used four tables for all the dishes the guests brought to the party. The tables were 2 8/10 meters long,

ddd [48]

Answer:

$2.83,2.8,2.59,2.48$

Step-by-step explanation:

Given:

Numbers are $\frac{28}{10}\,m,\,2.48\,m,\,\frac{259}{100}\,m,\,2.83 \,m$

To express: each length as a decimal number in order from greatest to least

Solution:

A number which consists of a whole number part and the fractional part separated by a decimal point is known as a decimal number.

$\frac{28}{10}=2.8\\ 2.48=2.48\\\frac{259}{100}=2.59\\ 2.83=2.83$

Numbers arranged in order from greatest to least: $2.83,2.8,2.59,2.48$

4 0

4 years ago

Find the general solution to: y^(4) - y =0

N76 [4]

Answer:

$y(t) = C_1 e^{t} + C_2 e^{-t} + D_1 \cos t + D_2 \sin t }$

Step-by-step explanation:

The equation is a linear differential equation: y⁽⁴⁾- y = 0

We assume the form of the solution y(t) is $y(t)=C_{1} e^{\alpha_{1} t} + C_{2} e^{\alpha_{2} t} + C_{3} e^{\alpha_{3} t} + C_{4} e^{\alpha_{4} t}$

where $$\alpha_{i}$ are the roots of the auxiliary equation.

So, use the auxiliary equation: $\alpha^4 + 0 \alpha^3 + 0 \alpha^2 + 0 \alpha -1 =0$ to find the roots; the values are : α₁ = 1, α₂ = -1, α₃ = i, α₄ = -i

Then inserting $$\alpha_{i}$ values in the assumed solution

⇒ $y(t)=C_1 e^{t} + C_2 e^{-t} + C_{3} e^{it} + C_{4} e^{-it}$

Also, because the last 2 terms have complex power, the solution can be written with cosine and sine terms:

Using the Euler's formula: $e^{ \pm i\theta } = \cos \theta \pm i\sin \theta$ , we can rewrite the solution as:

$$y(t) = C_{1} e^{t} + C_{2} e^{-t} + C_{3} e^{i t} + C_{4} e^{-i t}$ = $C_{1} e^{t} + C_{2} e^{-t} + C_{3} ( \cos t + i \sin t ) + C_{4} ( \cos t - i \sin t ) = C_{1} e^{t} + C_{2} e^{-t} + \cos t ( C_{3} + C_{4} ) + \sin t (i C_{3} - i C_{4} ) = C_{1} e^{t} + C_{2} e^{-t} + D_{1} \cos t +D_{2} \sin t$$