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3 years ago

4. k(x) × g(x) × f(x)

Mathematics

1 answer:

Vikentia [17]3 years ago

7 0

$k(x) \times f(x)$

$(2x + 5) \times (x - 1)$

$2 x(x - 1) + 5(x - 1)$

$({2 x }^{2} - 2x) + (5x - 5)$

${2x}^{2} - 3x - 5$

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$k(x) \times g(x) \times f(x)$

$g(x) \times (k(x) \times f(x))$

$({6x }^{2} - 4x + {3x}^{4} ) \times ( {2x}^{2} - 3x - 5)$

${6x}^{2} ( {2x}^{2} - 3x - 5) - 4x ( {2x}^{2} - 3x - 5) + {3x}^{4} ( {2x}^{2} - 3x - 5)$

$12 {x}^{4} - 18 {x}^{3} - 30 {x}^{2} - {8x}^{3} - 12 {x}^{2} - 20x + {6x}^{6} - 9x {}^{5} - {15x}^{4}$

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Reed made a lasagna for dinner. That night, he ate1/4

Snezhnost [94]

Answer: 11/12

Step-by-step explanation:

First find the LCM of 4 and 3(12). Then make the denominator of both fractions 12(3/12 and 8/12). Then add the fractions to get that they ate 11/2 of the lasagna.

Hope it helps <3

6 0

4 years ago

Find the slope of the line that passes through (-6, -2) and (-7, -2)

Jobisdone [24]

Answer:

$0$

Step-by-step explanation:

We could use the Slope Formula to find the answer, but there's an easier way of doing it. Notice that the y-coordinates of both points are the same. This means that the line that passes through these points is horizontal, which makes our answer $0$ because all horizontal lines have a slope of $0$ .

All horizontal lines have a slope of $0$ because the value of y never changes as the value of x changes, so if we were to input any two points on a horizontal line into the Slope Formula, our numerator would be $0$ (because anything subtracted from itself is $0$ ), leaving us with a result of $0$ regardless of the denominator since $0$ divided by anything is $0$ . Hope this helps!

6 0

3 years ago

Help would be greatly appreciated

Nesterboy [21]

I believe A 5 is correct

8 0

4 years ago

What is the value of x in the equetion 3/4(1/4x+8)-(1/2+2)-3/8(4-x)-1/4x?

ikadub [295]

3/4*1/4= 3/16 or 0.1875 (you times 3 by 1 and 4 by 4)
3/4*8= 6
first bracket = 3/16+ 6
1/2+2= 2 1/2 or 2.5
second bracket = -2.5
-3/8*4= -1.5
-3/8*x= -3/8x or -0.375x
third bracket= -0.375x
so

(0.1875+6) - (2.5) - (0.375x) - (0.25x)
collect like terms
6.2 - 2.5 = 3.7
0.375x - 0.25x = 0.125x
3.7-0.125x is your answer in decimal form

(3/16+6/1) - (2 1/2) - (3/8x-1/4x)

(6 3/16 - 2 1/2 )

(3 11/16)

3/8x-1/4x= 1/8x

final answer = 3 11/16 - 1/8x
-

6 0

3 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

8 0

4 years ago

4. k(x) × g(x) × f(x)​

4. k(x) × g(x) × f(x)