what is the slope of (-4,7) and (6,2)

Answer:

The solutions to the system of the equations by the elimination method will be:

[tex]x=2,\:z=-1,\:y=2[/tex]

Step-by-step explanation:

Given the system of the equations

[tex]2x\:+\:2y\:+z\:=\:7[/tex]

[tex]-x-\:y\:+z\:=\:-5[/tex]

[tex]x+3y-4z=12[/tex]

solving the system of the equations by the elimination method

[tex]\begin{bmatrix}2x+2y+z=7\\ -x-y+z=-5\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y+z=-5\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y+2z=-10[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ -2x-2y+2z=-10\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]-2x-2y+2z=-10[/tex]

[tex]+[/tex]

[tex]\underline{2x+2y+z=7}[/tex]

[tex]3z=-3[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]2x+6y-8z=24[/tex]

[tex]-[/tex]

[tex]\underline{2x+2y+z=7}[/tex]

[tex]4y-9z=17[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ 4y-9z=17\end{bmatrix}[/tex]

Rearranging the equations

[tex]\begin{bmatrix}2x+2y+z=7\\ 4y-9z=17\\ 3z=-3\end{bmatrix}[/tex]

solve [tex]3z=-3[/tex] for z:

[tex]z=-1[/tex]

[tex]\mathrm{For\:}4y-9z=17\mathrm{\:plug\:in\:}z=-1[/tex]

solve  [tex]4y-9\left(-1\right)=17[/tex] for y:

[tex]4y-9\left(-1\right)=17[/tex]

[tex]4y+9=17[/tex]

[tex]4y=8[/tex]

[tex]y=2[/tex]

[tex]\mathrm{For\:}2x+2y+z=7\mathrm{\:plug\:in\:}z=-1,\:y=2[/tex]

solve [tex]2x+2\cdot \:2-1=7[/tex] for x:

[tex]2x+2\cdot \:2-1=7[/tex]

[tex]2x+3=7[/tex]

[tex]2x=4[/tex]

[tex]x=2[/tex]

Therefore, the solutions to the system of the equations by the elimination method will be:

[tex]x=2,\:z=-1,\:y=2[/tex]

Answer:

0.3495.

Step-by-step explanation:

............................

Answers:

===================================================

Explanation:

The largest exponent term is what will go first. Then the term with the second largest goes next, and so on. We count the exponents down. The exponents will count down in this case as 5, 3, 1, 0

Note: -9 = -9*1 = -9m^0 and -11m = -11m^1

So that's why standard form is [tex]\frac{3}{5}m^5 + 7m^3 - 11m-9[/tex]

We can write this as (3/5)m^5 + 7m^3 - 11m - 9

------------------------

The leading term is the first term after getting everything in standard form. The coefficient of the leading term is the leading coefficient. So in this case, it would be 3/5.

The degree is the largest exponent. So the degree is 5. This is a quintic polynomial.

Answer:1222

Step-by-step explanation:

this is why this the answer

Answer:

Wht u need help on?

Step-by-step explanation:

No, because intergers don't include fractions.

Answer:

No

Step-by-step explanation:

Integers cannot have any fractions.

If you simplify this to get 8/1 and then 8, 8 is an integer

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:

have:

[tex]\frac{sin}{1-cot}+\frac{cos}{1-tan}\\\\=\frac{sin}{1-\frac{cos}{sin} }+\frac{cos}{1-\frac{sin}{cos} }\\\\=\frac{sin^{2} }{sin-cos}+\frac{cos^{2} }{cos-sin} \\\\=\frac{sin^{2} }{sin-cos}-\frac{cos^{2} }{cos-sin}\\\\=\frac{sin^{2}-cos^{2} }{sin - cos}\\\\=\frac{(sin-cos)(sin+cos)}{sin-cos}\\\\=sin+cos[/tex]

Step-by-step explanation:

The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14.

A pattern is a specific arrangement of numbers or any term which will follow a specific code or hint.

For example 2,4,6,8 in this, the difference between any two-term is only 2 and this will be its specific pattern.

As per the given pattern,29, 28, 26, 23, 19

29 - 28 = 1

28 - 26 = 2

26 - 23 = 3

23 - 19 = 4

Since the difference is increasing by 1

So the difference of 19 and the next number (x) will be 4 + 1 = 5.

19 - x = 5

x = 14

Hence "The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14".

For more information about the pattern,

brainly.com/question/14013716

#SPJ2

Answer:

B

Step-by-step explanation:

big line= divide

[tex]-2 1/4 to improper = -9/4\\/ -2/3[/tex]

so -9/4 / -2/3 or

B

Hope this helps :D

Answer:

B!

Step-by-step explanation:

if you calculate the original expression given you get [tex]3\frac{3}{8}[/tex] so i just went down the line calculating the rest of the equations and B was the one equivalent to the original expression! hope this helped.

Answer:

Step-by-step explanation:

n = -14

(x^3)(x^−17) =

x^(3 - 17) =

x^-14

Answer:

The most that they can spend in food is $239

Step-by-step explanation:

Their budget allows for less than $720 for both food and rent.

We are told that this budget allows

twice as much money for rent as

for food.

Thus,

Let x be amount of spent on food.

Thus, money spent on rent is 2x

Thus,

2x + x < 720

3x < 720

x < 720/3

x < 240

Since x should be less than 240,it means that the most that they can spend in food is $239

Answer:

where are the options

Step-by-step explanation:

Answer:

it cost $8, it was a swimsuit and he bought it

Step-by-step explanation:

Answer: See explanation

Step-by-step explanation:

1. Jun needs to review for the entrance exam with his remaining 5 3/5 hours fo the 7 subjects. how long will he spend for each subject?

The time spent on each subject will be the remaining hours left divided by 7. This will be:

= 5 3/5 ÷ 7

= 28/5 ÷ 7

= 28/5 × 1/7

= 28/35 = 4/5 hours

Since 60 minutes make one hour

4/5 hours = 4/5 × 60 = 48 minutes

2.Mar purchased 6 3/4 meters of wall paper for the boxes he will cover. if each box needs 2 1/4 meters, how many boxes will get covered?

This will be:

= 6 3/4 ÷ 2 1/4

= 27/4 ÷ 9/4

= 27/4 × 4/9

= 3 boxes

3.Joana has 12 1/4 meters of ribbon, joy has 3meters ribbon shorter than joana. they split the ribbon to have the an equal length of ribbon. how long is each.

Joana ribbon = 12 1/4 meters

Joy's ribbon = 12 1/4 - 3 = 9 1/4 meters

We then add both length of ribbons and divide by 2. This will be:

= 12 1/4 + 9 1/4

= 21 1/2 meters

= 21 1/2 meters ÷ 2

= 43/2 ÷ 2

= 43/2 × 1/2

= 43/4

= 10 3/4 meters

Answer:

you look at the numbers and add the other numbers.

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given that :

Dimension of rail road toy trains :

Length = 15 inches

Width = 4 inches

Scale = 2/9

Dimension of car on the long Island rail road :

Let length = x

Length = 2/9 of x = 15 inches

= 0.22222x = 15 inches

LENGTH = 67.5 inches

WIDTH = 2/9 of x = 4 inches

0.22222x = 4 inches

Width = 18

Width = 6 inches

Answer:

1) The equation that represents the number of slices of bacon to the number he will eat, assuming he eats 24 grams is;

3·x + 6·y = 24

2) The equation in slope and intercept form is;

y = 4 - 0.5·x

Step-by-step explanation:

1) The given parameters are;

The amount of protein Tommy wants to eat for breakfast = 24 grams

The quantity of protein in a slice of bacon = 3 grams

The quantity of protein in an egg = 6 grams

Let x represent the number of slices of bacon he will eat and let y represent the number of eggs he will eat

The equation that represents the number of slices of bacon to the number he will eat, assuming he eats 24 grams is given as follows;

3·x + 6·y = 24

2) The equation in slope and intercept form is given by making y the subject of the above equation as follows;

3·x + 6·y = 24

6·y = 24 - 3·x

y = 24/6 -1/2·x

y = 4 - 0.5·x, which is in the slope and intercept form y = m·x + c

Where;

m = The slope = -0.5

c = The y-intercept = 4

Answer:

1. -32

2. -46

Step-by-step explanation:

1. When muliplying a number with a postive number and a negative number the answer will be a negative number

2. When you have a negative number and when  you have a postive number and you are subtracting your answer has to be a negative answer

Answer:

cvgbnm,m

Step-by-step explanation:

bxxcvn vn

Answer: 600

Step-by-step explanation:

I just took the test

Answer:

Hi it’s me

Step-by-step explanation:

7is the first one

10 is the second

Nvm you already got the answer cause yo asked this 2 weeks ago anyways hi

Answer:

A ( SAS )

Step-by-step explanation:

Answer:

x < 1

Step by step explanation:

10 + 4x < 14

4x < 14 -10

4x < 4

x < 1

When you graph it on a number line, draw a line that extends from 1 towards the left side (since x is less that 1) and use unshaded circles.

Answer:

to find the line you must find x and y

Step-by-step explanation:

[tex]\left[\begin{array}{ccc}x&-6&0\\y&0&3\end{array}\right][/tex]

graph the line using slope and y intercept

slop:1/2

y-intercept:3

find the values of mx and b

[tex]m=\frac{1}{2}\\b=3[/tex]

find x intercept

[tex]0=\frac{x}{2}+3[/tex]

re wright the equasion

[tex]\frac{x}{2}+3=0[/tex]

subtract 3

[tex]\frac{x}{2} +3-3=0-3[/tex]

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

multiply by 2

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

x intercept = -6

now find y intercept

[tex]y=\frac{0}{2}+3[/tex]

divide by 2

[tex]y=0+3[/tex]

add 0 and 3

y=3

create a table of x and y [tex]\left[\begin{array}{ccc}x&-6&0\\y&0&3\end{array}\right][/tex]

graph the line using y-intercept and slope

so your answer is (-6,3)

Answer:

A

Step-by-step explanation:

Answer:

Step-by-step explanation:

Plug in the stated values for x

(a) 2(1)^4 + 3(1)^3 - (1)^3 + 2(1) - 3

   2 + 3  - 1 + 2 - 3 = 1 + 2 = 3

  f(1) = 3

(b) 2(-1)^4 + 3(-1)^3 - (-1)^2 + 2(-1) - 3

2 * 1 + 3(-1) - 1 + 2(-1) - 3

2 + -3 -1 -2 - 3 = 2 - 9 = -7

f(-1) = -7

(c) 2(0) + 3(0) - 0 + 0 - 3

0 - 3 = -3

f(0) = -3

d) 2(-1/2)^4 + 3(-1/2)^3 - 2(-1/2) - 3

1/8 - 3/8 - (-1/2)^2 - 2 * 1/2 - 3

16/4 - 1/4 - 1/4 = -16-1-1/4 = -18/4

f(-1/2) = -9/2

Answer:

Area of ΔDEF is 2.25 times the area of ΔABC.

Step-by-step explanation:

Since, ΔDEF is the enlarged form of ΔABC,

Scale factor = [tex]\frac{\text{Side length of triangle ABC}}{\text{Corresponding side length of image triangle}}[/tex]

                    = [tex]\frac{15}{10}[/tex]

                    = 1.5

Since, ratio of the areas of image and preimage triangles = (Scale factor)²

[tex]\frac{\text{Area of image triangle}}{\text{Area of triangle ABC}}[/tex] = (Scale factor)²

[tex]\frac{\text{Area of image triangle}}{\text{Area of triangle ABC}}=(1.5)^2[/tex]

Area of image ΔDEF = 2.25(Area of ΔABC)

Therefore, area of ΔDEF is 2.25 times the area of ΔABC.