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3 years ago

The function notation f (x) means "f multiplied by x." True False

Mathematics

2 answers:

34kurt3 years ago

8 0

False f(x) is another way to say y is equal to

Digiron [165]3 years ago

6 0

Your answer is True!!

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I need help asap please

rjkz [21]

Answer:

$x = -\frac{3}{2}$ or $x = 1$

Step-by-step explanation:

Using the zero product property, first step is to set the given equation, $2x^2 + x - 1 = 2$ , to zero. Then factorise the left side.

Thus,

$2x^2 + x - 1 = 2$

Subtract 2 from both sides

$2x^2 + x - 1 - 2 = 2 - 2$

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

Factorise the left side

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

$x(2x + 3) - 1(2x + 3) = 0$

$(x - 1)(2x + 3) = 0$

Find the solution

$x - 1 = 0$

$2x + 3 = 0$

$x = 1$

$2x + 3 = 0$

$2x = -3$

$x = -\frac{3}{2}$

The answer is: $x = 1$ or $x = -\frac{3}{2}$

3 0

2 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.