All categories

4 years ago

What is the answer to this math problem, g=c+x

Mathematics

1 answer:

barxatty [35]4 years ago

3 0

Step-by-step explanation:

Q: Solve for x

g = c + x

g - c = x

You might be interested in

Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var

Natalija [7]

Answer:

$\displaystyle y=5+\frac{x}{10}$

$\displaystyle y=10+\frac{(x-75)}{20}$

Step-by-step explanation:

<u>Linearization</u>

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

$y=f(x)=f(a)+f'(a)(x-a)$

Let's find the linearization for the function

$y=\sqrt{25+x}$

at (0,5) and (75,10)

Computing f'(x)

$\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}$

At x=0:

$\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}$

We find f(0)

$f(0)=\sqrt{25+0}=5$

Thus the linearization is

$\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x$

$\displaystyle y=5+\frac{x}{10}$

Now at x=75:

$\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}$

We find f(75)

$f(75)=\sqrt{25+75}=10$

Thus the linearization is

$\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)$

$\displaystyle y=10+\frac{(x-75)}{20}$

5 0

3 years ago

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

$y=(x-1)^2-4$

$(y+4)=(x-1)^2$

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.