Ask question

Ask question

All categories

chubhunter [2.5K]

2 years ago

12

A plumber charges $55.00 for a service call plus $35.00 per hour. If the total bill was $160.00 . Select an equation , with x re

presenting the number of hours that can be use to determine how many hours the plumber was there.
A) 55x + 35 = 160
B) 35x + 160 = 55
C) 35x + 55 = 160
D) 160x +55 = 35.00

2 answers:

Vladimir [108]2 years ago

7 0

Answer:

its C that's definitely correct

Anestetic [448]2 years ago

7 0

Pretty sure its C if not I’m very sorry

You might be interested in

Help in math ...........................................

Sauron [17]

Your answer would be B.

The triangles would be congruent, because the lines GH and FI are congruent and the angle H and I are the same, and since the line HI is shared by the two, it would then be SAS.

7 0

2 years ago

Read 2 more answers

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

Solve the following equation : 6x =12

densk [106]

X=2 here very simple

8 0

3 years ago

2x -3y =2<br><br> x = 6y - 5

never [62]

2(6y - 5) - 3y = 2
12y - 10 - 3y = 2
9y - 10 = 2
9y = 12
y = 4/3, or 1 1/3
2x - 3(4/3) = 2
2x - 12/3 = 2
2x - 4 = 2
2x = 6
x = 3

7 0

3 years ago

A store was selling lawn mowers for $69. The repair

Natali5045456 [20]

Answer:

$84.87

Step-by-step explanation:

The mover costs $69

23% of 69 is $15.87

add the two together to get $84.87

(or just multiply 69 by 1.23)

5 0

3 years ago

Read 2 more answers