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2 years ago

Let f(x)=x2-3x + 9. Find f(-1).

Mathematics

1 answer:

netineya [11]2 years ago

3 0

X^2 - 3x + 9
(-1)^2 - 3(-1) + 9
1 - 3(-1) + 9
1 + 3 + 9
13

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Write an equation of the line that passes through the points (-2,-3) and (1-3)

Ronch [10]

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-3+3}{1+2}\implies \cfrac{0}{3}\implies 0$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{0}[x-\stackrel{x_1}{(-2)}] \\\\\\ y+3=0\implies y=-3$

7 0

3 years ago

Read 2 more answers

What is the product? (6r-1)(- Br-3)

Andru [333]

Answer:

$-6Br^2-18r+Br+3$

Step-by-step explanation:

we have the expression:

$(6r-1)(- Br-3)$

We need to multiply the two terms in the first parenthesis by the two terms in the second parenthesis:

$(6r)(-Br)+(6r)(-3)+(-1)(-Br)+(-1)(-3)$

we solve each term taking into account the law of signs

(-)(-)=+

and

(+)(-)=-

also for the first term we use that $r*r=r^2$

thus we have:

$-6Br^2-18r+Br+3$

it can be simplified a little more if we factor an r of the second and third term:

$-6Br^2+r(-18+B)+3$

4 0

3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Recall the scenario about Eric's weekly wages in the lesson practice section. Eric's boss have been very impressed with his work

Alona [7]

Answer:

$1)\quad f(x)=\bigg\{\begin{array}{ll}12x&0\leq x$

2) D: x = [0, 24]

3) R: y = [0, 384]

4) see graph

Step-by-step explanation:

Eric's regular wage is $12 per hour for all hours less than 9 hours.

The minimum number of hours Eric can work each day is 0.

f(x) = 12x for 0 ≤ x < 9

Eric's overtime wage is $18 per hour for 9 hours and greater.

The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).

f(x) = 18(x - 8) + 12(8)

= 18x - 144 + 96

= 18x - 48 for 9 ≤ x ≤ 24

The daily wage where x represents the number of hours worked can be displayed in function format as follows:

$f(x)=\bigg\{\begin{array}{ll}12x&0\leq x$

2) Domain represents the x-values (number of hours Eric can work).

The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.

D: 0 ≤ x ≤ 24 → D: x = [0, 24]

3) Range represents the y-values (wage Eric will earn).

Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.

The minimum hours he can work in one day is 0.

f(x) = 12x

f(0) = 12(0)

= 0

The maximum hours he can work in one day is 24 (although unlikely, it is theoretically possible).

f(x) = 18x - 48

f(24) = 18(24) - 48

= 432 - 48

= 384

D: 0 ≤ y ≤ 384 → D: x = [0, 384]

4) see graph.

Notice that there is an open dot at x = 9 for f(x) = 12x

and a closed dot at x = 9 for f(x) = 18x - 48

4 0

3 years ago

Curtis plays a trivia game he earns 2 points for each correct answer and loses 0.25 points for each incorrect answer he get 21 c

Maslowich

Answer:

Step-by-step explanation:

Curtis is playing a trivia game.

Number of points earned for each correct answer = 2

Number of points lost for each correct answer = 0.25

Given that:

Number of correct answers given by Curtis = 21

Number of wrong answers given by Curtis = 4

To find:

Total score of the game by Curtis?

Solution:

First of all, let us find the points earned by correct answers.

Points earned by correct answers = Number of correct answers multiplied by Points earned on one correct answer.

Points earned by correct answers = 21 $\times$ 2 = 42

Now, let us find the points to be lost by wrong answers.

Points lost by wrong answers = Number of wrong answers multiplied by Points lost on one correct answer.

Points lost by wrong answers = 4 $\times$ 0.25 = 1

The score of the game can be calculated by subtracting the points lost from the points earned.

Score = 42 - 1 = 41

3 0

2 years ago