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3 years ago

Solve 6x+2=2x-2 these sort of equations I cannot understand

Mathematics

1 answer:

ss7ja [257]3 years ago

8 0

6x+2=2x-2
6x+2-2=2x-2-2
6x=2x-4
6x-2x=2x-4-2x
4x=-4
4x/4= -4/4
x= -1
hope this helps :)

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If 1m = 3.3 ft, how many cubic feet is the Kuroshio Sea tank? Volume is 9,450. Show your calculations!

Naddika [18.5K]

Answer:

339604.65 cubic feet

Step-by-step explanation:

We are given that 1 m is equal to 3.3 ft.

1 m = 3.3 ft

This means that if we find the volume of a cube of 1 m, we have its volume as:

$1^3 = 1 m^3$

In the same way, the volume of a cube of 3.3 ft (1 m) is:

$3.3^3 = 35.937 ft^3$

This means that:

$1m^3 = 35.937 ft^3$

The Kuroshio sea tank has volume of $9450 m^3$ . Its volume in cubic feet is therefore:

$1m^3 = 35.937 ft^3\\\\9450 m^3 = 9450 * 35.937 = 339604.65 ft^3$

The volume of the tank is 339604.65 cubic feet.

3 0

3 years ago

Sovle: 10 -2x = -30 Please use subsitution Show your steps to solve the equation

Margaret [11]

10 - 2x = -30
-2x = -40
2x = 40
x = 20

7 0

3 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

write an expression that has a value of 100 and follows these restrictions. (1) uses each of the four operations at most once. (

posledela

33 * 3 + 3/3 = 100

I do not believe u have to use all of the operations because it says " at most once ".

5 0

3 years ago

Read 2 more answers

Rewrite the following equation in logarithmic form.

zloy xaker [14]

Answer:

log2(0.25) = -2

512 = 8^3

Step-by-step explanation:

Use of formula: loga (b) = c ⇔ b = a^c

Rewrite the following equation in logarithmic form. 0.25 = 2^-2

log2(0.25) = -2

Rewrite the following equation in exponential form. log8 (512) = 3

512 = 8^3

3 0

3 years ago