What is the value of x and what does ACB equal to? Please answer as soon as possible! Thanks for your support!

Answer:

6A7C8B

Step-by-step explanation:

Add them in and try to solve for 7 and 6, for 8. I usually put all number on one side and all x on other side.

Step-by-step explanation:

the regular form of r=8

Answer:

a= (Z + 2b)/24

Step-by-step explanation:

given:

Z = 24a - 2b

to solve for a, we will try to move "a" to one side of the equation and all other terms to the other side:

Z = 24a - 2b    (add 2b to both sides)

Z + 2b = 24a -2b +2b

Z + 2b = 24a  (switch sides)

24a = Z + 2b (divide both sides by 24)

a/24 = (Z + 2b)/24

a= (Z + 2b)/24

Answer:

16.5 pages

Step-by-step explanation:

Number of pages Ingrid types = 3

Number of pages Tanya types = 4.5

The proportion to find how many pages Tanya would have typed (x), if Ingrid typed 11 pages would be as follows:

3 : 4.5 = 11 : x

Thus:

[tex] \frac{3}{4.5} = \frac{11}{x} [/tex]

Cross multiply

[tex] 3*x = 11*4.5 [/tex]

[tex] 3x = 49.5 [/tex]

Divide both sides by 3

x = 16.5 pages

Answer:           0.875

Step-by-step explanation:

Answer:

Ich weiß nicht =13

die Antwort -2 -4 (6p-5)

Step-by-step explanation:

Answer:

5/2

Step-by-step explanation:

25 and 10 so 25/10 or 5/2 or 2 1/2

Answer:

0.0425

Step-by-step explanation:

Answer:

0.0425

Step-by-step explanation:

hope this helped ^^

Answer:

Step-by-step explanation:

1. use the slope formula --> (y2-y1)/(x2-x1)

2. (8 - 4)/(13 -(-3))

3. 4/16

4. m = 1/4

Answer:

-1.48148148

Step-by-step explanation:

-7 x -8 = 56

-7 x -4a = 28a

25-56=-31

27a/-31

a=-1.48148148

Answer:

the mean of which the team won is 8.75

The mean of which the team lost is 11.5

Step-by-step explanation:

add all the numbers of which the team won by then divide the sum by the number of numbers there are

example:

7+8+11+9=35/4=8.75

Answer:

Step-by-step explanation:

Rule for the dilation of a point by a scale factor 'k',

(x, y) → (kx, ky)

If we impose this rule in this problem,

k = [tex]\frac{\text{Distance of A' from P}}{\text{Distance of A from P}}[/tex]

1). If k = 3

   Therefore, Distance of A' from P = 3(Distance of A from P)

   And point A' will be on the third circle.

2). If k = 2

   Distance of B' from P = 2(Distance of B from P)

   Since, B is on circle 2, B' will be on circle 4.

Now we can plot these points A' and B' on the graph.

After the dilation point A becomes A' and it is at a distance of 3 units from point P and after the dilation point B becomes B' and it is at a distance of 2 units from point P.

1)

Given :

Dilate A using P as the center of dilation and a scale factor of 3.

Let the coordinates of point A be (x,y) then after dilation point A becomes A'(3x , 3y). So, the distance of the point A' from the point P is 3 units

2)

Given :

Dilate Busing P as the center of dilation and a scale factor of 2.

Let the coordinates of point B be (x',y') then after dilation point B becomes B'(2x' , 2y'). So, the distance of the point B' from the point P is 2 units

For more information, refer to the link given below:

https://brainly.com/question/2856466

Answer:

Let's define the symbols:

x <  y

means that x is strictly less than y.

x ≤ y

means that x is less than or equal to y.

(both of these signs can be in the other direction, take that in mind).

A) " a is always less than another number b"

Here we should have:

a < b.

a is always less than the other number b.

This is not in the options, so i suppose that there is a mistake in the question.

B) The expression "above" is not shown here, so i will just writhe the correspondent expression for each option, and you can see which one matches with the expression above.

a) "The product between 3 and a number is no less than 9"

3*n ≥ 9.

b) "The product between 3 and a number is greater than 9."

3*n > 9

c) "The product between 3 and a number is no more than 9"

3*n ≤ 9

d) "The product between 3 and a number is less than 9"

3*n < 9

Answer:

the first questions answer is yes the seconds is no

HAVE A NICE DAY!!!

Answer:

A. -x² + 2x + 8

General Formulas and Concepts:

Algebra I

Step-by-step explanation:

Step 1: Define

f(x) = 2x + 1

g(x) = x² - 7

(f - g)(x) is f(x) - g(x)

Step 2: Solve

Answer:

(1,1)

Step-by-step explanation:

Answer:

niwebfasfbjsdfb

Step-by-step explanation:

jsdfhsdbfjhadsjbdsfhb

Answer:

The solution is too long. So, I included them in the explanation

Step-by-step explanation:

This question has missing details. However, I've corrected each question before solving them

Required: Determine the inverse

1:

[tex]f(x) = 25x - 18[/tex]

Replace f(x) with y

[tex]y = 25x - 18[/tex]

Swap y & x

[tex]x = 25y - 18[/tex]

[tex]x + 18 = 25y - 18 + 18[/tex]

[tex]x + 18 = 25y[/tex]

Divide through by 25

[tex]\frac{x + 18}{25} = y[/tex]

[tex]y = \frac{x + 18}{25}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{x + 18}{25}[/tex]

2. [tex]g(x) = \frac{12x - 1}{7}[/tex]

Replace g(x) with y

[tex]y = \frac{12x - 1}{7}[/tex]

Swap y & x

[tex]x = \frac{12y - 1}{7}[/tex]

[tex]7x = 12y - 1[/tex]

Add 1 to both sides

[tex]7x +1 = 12y - 1 + 1[/tex]

[tex]7x +1 = 12y[/tex]

Make y the subject

[tex]y = \frac{7x + 1}{12}[/tex]

[tex]g'(x) = \frac{7x + 1}{12}[/tex]

3: [tex]h(x) = -\frac{9x}{4} - \frac{1}{3}[/tex]

Replace h(x) with y

[tex]y = -\frac{9x}{4} - \frac{1}{3}[/tex]

Swap y & x

[tex]x = -\frac{9y}{4} - \frac{1}{3}[/tex]

Add [tex]\frac{1}{3}[/tex] to both sides

[tex]x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}[/tex]

[tex]x + \frac{1}{3}= -\frac{9y}{4}[/tex]

Multiply through by -4

[tex]-4(x + \frac{1}{3})= -4(-\frac{9y}{4})[/tex]

[tex]-4x - \frac{4}{3}= 9y[/tex]

Divide through by 9

[tex](-4x - \frac{4}{3})/9= y[/tex]

[tex]-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y[/tex]

[tex]\frac{-4x}{9} - \frac{4}{27}= y[/tex]

[tex]y = \frac{-4x}{9} - \frac{4}{27}[/tex]

[tex]h'(x) = \frac{-4x}{9} - \frac{4}{27}[/tex]

4:

[tex]f(x) = x^9[/tex]

Replace f(x) with y

[tex]y = x^9[/tex]

Swap y with x

[tex]x = y^9[/tex]

Take 9th root

[tex]x^{\frac{1}{9}} = y[/tex]

[tex]y = x^{\frac{1}{9}}[/tex]

Replace y with f'(x)

[tex]f'(x) = x^{\frac{1}{9}}[/tex]

5:

[tex]f(a) = a^3 + 8[/tex]

Replace f(a) with y

[tex]y = a^3 + 8[/tex]

Swap a with y

[tex]a = y^3 + 8[/tex]

Subtract 8

[tex]a - 8 = y^3 + 8 - 8[/tex]

[tex]a - 8 = y^3[/tex]

Take cube root

[tex]\sqrt[3]{a-8} = y[/tex]

[tex]y = \sqrt[3]{a-8}[/tex]

Replace y with f'(a)

[tex]f'(a) = \sqrt[3]{a-8}[/tex]

6:

[tex]g(a) = a^2 + 8a- 7[/tex]

Replace g(a) with y

[tex]y = a^2 + 8a - 7[/tex]

Swap positions of y and a

[tex]a = y^2 + 8y - 7[/tex]

[tex]y^2 + 8y - 7 - a = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 8[/tex]; [tex]c = -7 - a[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-8 \±\sqrt{8^2 - 4 * 1 * (-7-a)}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{64 + 28 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 }[/tex]

Factorize

[tex]y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }[/tex]

[tex]y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }[/tex]

[tex]y = -4 \±\sqrt{(23 + a)}[/tex]

[tex]g'(a) = -4 \±\sqrt{(23 + a)}[/tex]

7:

[tex]f(b) = (b + 6)(b - 2)[/tex]

Replace f(b) with y

[tex]y = (b + 6)(b - 2)[/tex]

Swap y and b

[tex]b = (y + 6)(y - 2)[/tex]

Open Brackets

[tex]b = y^2 + 6y - 2y - 12[/tex]

[tex]b = y^2 + 4y - 12[/tex]

[tex]y^2 + 4y - 12 - b = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 4[/tex]; [tex]c = -12 - b[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-4\±\sqrt{4^2 - 4 * 1 * (-12-b)}}{2*1}[/tex]

[tex]y = \frac{-4\±\sqrt{4^2 - 4 *(-12-b)}}{2}[/tex]

Factorize:

[tex]y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{16+b}}{2}[/tex]

[tex]y = -2\±\sqrt{16+b}[/tex]

Replace y with f'(b)

[tex]f'(b) = -2\±\sqrt{16+b}[/tex]

8:

[tex]h(x) = \frac{2x+17}{3x+1}[/tex]

Replace h(x) with y

[tex]y = \frac{2x+17}{3x+1}[/tex]

Swap x and y

[tex]x = \frac{2y+17}{3y+1}[/tex]

Cross Multiply

[tex](3y + 1)x = 2y + 17[/tex]

[tex]3yx + x = 2y + 17[/tex]

Subtract x from both sides:

[tex]3yx + x -x= 2y + 17-x[/tex]

[tex]3yx = 2y + 17-x[/tex]

Subtract 2y from both sides

[tex]3yx-2y =17-x[/tex]

Factorize:

[tex]y(3x-2) =17-x[/tex]

Make y the subject

[tex]y = \frac{17 - x}{3x - 2}[/tex]

Replace y with h'(x)

[tex]h'(x) = \frac{17 - x}{3x - 2}[/tex]

9:

[tex]h(c) = \sqrt{2c + 2}[/tex]

Replace h(c) with y

[tex]y = \sqrt{2c + 2}[/tex]

Swap positions of y and c

[tex]c = \sqrt{2y + 2}[/tex]

Square both sides

[tex]c^2 = 2y + 2[/tex]

Subtract 2 from both sides

[tex]c^2 - 2= 2y[/tex]

Make y the subject

[tex]y = \frac{c^2 - 2}{2}[/tex]

[tex]h'(c) = \frac{c^2 - 2}{2}[/tex]

10:

[tex]f(x) = \frac{x + 10}{9x - 1}[/tex]

Replace f(x) with y

[tex]y = \frac{x + 10}{9x - 1}[/tex]

Swap positions of x and y

[tex]x = \frac{y + 10}{9y - 1}[/tex]

Cross Multiply

[tex]x(9y - 1) = y + 10[/tex]

[tex]9xy - x = y + 10[/tex]

Subtract y from both sides

[tex]9xy - y - x = y - y+ 10[/tex]

[tex]9xy - y - x = 10[/tex]

Add x to both sides

[tex]9xy - y - x + x= 10 + x[/tex]

[tex]9xy - y = 10 + x[/tex]

Factorize

[tex]y(9x - 1) = 10 + x[/tex]

Make y the subject

[tex]y = \frac{10 + x}{9x - 1}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{10 + x}{9x -1}[/tex]

Answer:

The y intercept is (0, 8). The equation of the function is f(x) = 10x + 8

Step-by-step explanation:

Given that after 1 hour of work, Jack will make 18 dollars and the slope is 10, subtract 10 dollars from 18 dollars to find his set amount, or y-intercept.

18 - 10 = 8 dollars.

Now that we have the m and b values, we can create our equation.

The equation in standard form is f(x) = mx +b, where m = the slope and b = the y-intercept. Plug our values in and the equation will be f(x) = 10x + 8. You can test if this is correct by plugging in x values from the table and seeing if your calculated value correctly corresponds to the given y value in the table.

Answer:

It's 576 if you mean 64 times 9!

Step-by-step explanation:

Answer:

R= - 95/7

Step-by-step explanation:

Answer:

C. [tex] d = \frac{at^2}{2} [/tex]

Step-by-step explanation:

Given:

[tex] t = \sqrt{\frac{2d}{a}} [/tex]

Required:

Make d the subject of the formula.

SOLUTION:

[tex] t = \sqrt{\frac{2d}{a}} [/tex]

Square both sides

[tex] t^2 = (\sqrt{\frac{2d}{a}})^2 [/tex]

[tex] t^2 = \frac{2d}{a} [/tex]

Multiply both sides by a

[tex] a*t^2 = \frac{2d}{a}*a [/tex]

[tex] at^2 = 2d [/tex]

Divide both sides by 2

[tex] \frac{at^2}{2} = \frac{2d}{2} [/tex]

[tex] \frac{at^2}{2} = d [/tex]

[tex] d = \frac{at^2}{2} [/tex]

Answer:

20x^5

Step-by-step explanation:

You just multiply 5 by 4 and get 20 and when multiplying exponents you just add them together so you get 20x^5

Answer:X=3

y= -4

Step-by-step explanation:

If we're solving for x or y, then the intercept we're not solving for is equal to zero. Solving for y means x=0, solving for x means y=0.

so solving for x, 4x=12--> x=3, since y=0 in this scenario.

Solving for y, x=0, so -3y=12--> y=(-4)

9514 1404 393

Answer:

  see below for a table

Step-by-step explanation:

You can easily identify intercept points by solving for one variable when the other is zero.

x-intercept (y=0)

  4x = 12

  x = 12/4 = 3

y-intercept (x=0)

  -3y = 12

  y = 12/-3 = -4

So, the coordinate pairs of these points are ...

  (3, 0) and (0, -4)

__

If you solve for y, you get ...

  -3y = 12 -4x

  y = (4/3)x -4 . . . . . divide by -3

This suggests that x being a multiple of 3 will be useful for finding integer values of y.

We already have points for x=0 and x=3. We can add points for x=6 and x=-3.

  y = 4/3(6) -4 = 4

  y = 4/3(-3) -4 = -8

So, the additional points are ...

  (6, 4) and (-3, -8)