Calculus

x2 + 2x + 4 can be expressed in the form ax2 + bx + c, where a=1, b=2, c=4.

So you will be looking for two numbers that add to 2 [b], and multiply to 4 [ac]. Clearly, there are no two real numbers that can do this. Therefore the expression above cannot be factorised.

If the x=4, then its equivelent to 12 + 4, which equals 16.

The logarithm of 1 to the base 1 is indeterminate. The logarithm of a number x to the base a is a number y, such that ay = x. The most common base a is 10, or the natural base a is e (2.718281828...). It is invalid to think of logarithms base 1, because 1 to the power of anything is still 1.

shut up and do your hw

the answer is x = 0

Add the two equations: x + y - x + 2y = 8 + 7 ie 3y = 15 so y = 5 making x = 3.

My opinion: this is a very bad idea

The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.

15x2 - 30x -720 = 15(x - 8)(x + 6).

Here are possibilities of what you mean:

a: (3-2)(x-2)=10

Solution:

(3-2)(x-2)=10

1(x-2)=10

x-2=10/1

x=10+2=12

b: 3-2(x-2)=10

Solution:

3-2(x-2)=10

3-10=2(x-2)

-7=2x-4

-3=2x

x=-3/2=-1.5

• y=0.5x is an equation that cannot be solved unless you have a known value for either y or x.

-9-2x+x = -13

-2x+x = -13+9

-x = -4

x = 4

The idea is to use the addition/subtraction property. In other words, take the derivative of 5x, take the derivative of 1, and subtract the results.

4

penerapan integral dalam bidang teknik

Add like terms:

(4y - y) + (23 - 6)

= 3y + 17

No. The so-called "natural" logarithms have a base of ' e ', and you can find

the log of any positive number to any base you like.

NO

x+3y=9

First I would solve for y

3y=-x+9

Now divide by 3

y=-x/3 + 3

This is what is called slope - intercept form.

The y-intercept is 3. So, you have your first point (0,3)

now to make this easy - choose any number that is a multiple of 3 (other than 0). How about 3.

Plug x=3 into the equation. y=-3/3+3 => y=-1+3 => y=2

So (3,2) is another point on this equation.

Now that you have 2 points you can draw the graph since it is linear. Just draw a straight line from (0,3) to (3,2) and continue drawing to the right and to the left.

(x-5)(x+3)

x^2+y^4=10x+7 @ (1,-2)

Rearrange to isolate y.

y=(-x^2+10x+7)^(1/4)

Take derivative.

y'=(1/4)*((-x^2+10x+7)^(-3/4))*(-2x+10)

y'=(-2x+10)/4*(-x^2+10x+7)^(3/4)

Sub in x=1.

y'=8/2*(16)^(3/4)

y'=2/8

y'=1/4

y'=m=slope=1/4

Sub m and (1,-2) into linear equation (tangent line).

m*(x-x1)=y-y2

(1/4)(x-1)=y+2

Isolate y.

Equation of tangent line: y=(1/4)x-(9/4).

The third question is unclear.

X = 7

9x - 2x = 49

x = 6

So y = 9*6 + 4 = 54 + 4 = 58

The ordered pair is (6, 58)

Given:

x + y = 5

2x + y = 6

We can subtract the second equation from the first to get:

-x = -1

x = 1.

From there, substituting back in to the first equation yields:

1 + y = 5

y = 4.

The solution is (1, 4).