What is the absolute value of -2/3??? PLEASE HELP!

Mathematics

1 answer:

d1i1m1o1n [39]4 years ago

7 0

Answer:

2/3

Step-by-step explanation:

The absolute value of a negative number will switch to a positive when the indicators are removed.

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write the equation in standard form of the function with x intercepts at (4,0) and (-2,0) that has been reflected across the x a

Anastaziya [24]

answer

$-3x^{2} +6x + 24$

explanation

since there are 2 x-intercepts, the lowest degree the function could possibly have is 2, so this could be the graph of a parabola

since our zeros are at x = 4 and x = -2, our factors are (x - 4) and (x + 2)

our equation so far is

(x - 4)(x + 2)

= $x^2 -4x + 2x - 8$

= $x^{2} -2x - 8$

since the function is reflected across the x-axis, multiply the equation by -1

$-(x^{2} -2x - 8)$

= $-x^{2} +2x + 8$

since the function has a vertical stretch of 3, multiply the equation by 3

$3(-x^{2} +2x + 8)$

= $-3x^{2} +6x + 24$

3 0

3 years ago

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$