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3 years ago

This system has one solution.

Mathematics

2 answers:

lana [24]3 years ago

4 0

Answer:

y = 7

Step-by-step explanation:

VMariaS [17]3 years ago

3 0

Answer:

y=7

Step-by-step explanation:

The given system is

$y=x^2-3x+7$

and

$y=3x-2$

We equate the two equations to find their point of intersection.

$x^2-3x+7=3x-2$

$x^2-3x-3x+7+2=0$

$x^2-6x+9=0$

$(x-3)^2=0$

$x-3=0$

$x=3$

The y-coorinate is

$y=3(3)-2$

$y=9-2=7$

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What are the values for matrix A?: 1 2 3 4 1 3 4 5 6 7 8 Noo 4 1 0 0 9 6 7 8 10 12 14 15 16 13 14 14 15 16

Gala2k [10]

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7 0

3 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.