PLEASE HELP!!!!!! I AM BEING TIMED!!!!
A garden hose fills a 2 gallon bucket in five seconds the number of gallons G is the proportional to the number of seconds T that the water is running select all the equations that represent the relationship between G and T

Answer:

2x-5y=-3

Slope=  2.000 /0.800

​x-intercept=-3/2=-1.50000

y-intercept=3/5=0.60000

y=16-4x

Slope=-8.0000/2.0000=-4.000

x-intercept=16/4=4

y-intercept=6/1=16.00000

Step-by-step explanation:

Answer:

-74+3i

Step-by-step explanation:

Answer:

Step-by-step explanation:

(5 + 3i) + (2 - 81)

5 + 3i + 2 - 81

74 + 3i

Answer:

-1.333333

Step-by-step explanation:

:)

have a very nice day

Answer:

I believe the answer is 42 minutes :)

Answer:

Step-by-step explanation:

500×30=15000

51.1 and 64.9

Step-by-step explanation:

Answer: 55.1 and 64.9

On edge.

Step-by-step explanation:

Answer:

= −6x+10

Step-by-step explanation:

Answer:

-6x+10

Step-by-step explanation:

Multiply the 2 by each figure in the parenthesis. So, it becomes the sum of (2)-3x and 2(5), which is -6x+10.

Answer:

7776

Step-by-step explanation:

Answer:

0 cavities

minimum is 0 cavities. there are people having 0 cavities. see the graph.

hope it helps!

Answer:

0

Step-by-step explanation:

The minimum is the smallest value in the dataset. The smallest number of cavities is 0.

please give thanks by clicking the heart button! :)

Answer:

a. [tex](x - 3)^2 + 16[/tex]

b. [tex]8(x -7)^2[/tex]

c. [tex](a^2 - 1)(7x - 6)[/tex] or [tex](a+1)(a-1)(7x-6)[/tex]

d. [tex](x^2-4)(x^2+3)[/tex] or [tex](x-2)(x+2)(x^2+3)[/tex]

e. [tex](a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})[/tex]

Step-by-step explanation:

[tex]a.\ (x + 1)^2 - 8(x - 1) + 16[/tex]

Expand

[tex](x + 1)(x + 1) - 8(x - 1) + 16[/tex]

Open brackets

[tex]x^2 + x + x + 1 - 8x + 8 + 16[/tex]

[tex]x^2 + 2x + 1 - 8x + 24[/tex]

Collect Like Terms

[tex]x^2 + 2x - 8x+ 1 + 24[/tex]

[tex]x^2 - 6x+ 25[/tex]

Express 25 as 9 + 16

[tex]x^2 - 6x+ 9 + 16[/tex]

Factorize:

[tex]x^2 - 3x - 3x + 9 + 16[/tex]

[tex]x(x -3)-3(x - 3) + 16[/tex]

[tex](x - 3)(x - 3) + 16[/tex]

[tex](x - 3)^2 + 16[/tex]

[tex]b.\ 8(x - 3)^2 - 64(x-3) + 128[/tex]

Expand

[tex]8(x - 3)(x - 3) - 64(x-3) + 128[/tex]

[tex]8(x^2 - 6x+ 9) - 64(x-3) + 128[/tex]

Open Brackets

[tex]8x^2 - 48x+ 72 - 64x+192 + 128[/tex]

Collect Like Terms

[tex]8x^2 - 48x - 64x+192 + 128+ 72[/tex]

[tex]8x^2 -112x+392[/tex]

Factorize

[tex]8(x^2 -14x+49)[/tex]

Expand the expression in bracket

[tex]8(x^2 -7x-7x+49)[/tex]

Factorize:

[tex]8(x(x -7)-7(x-7))[/tex]

[tex]8((x -7)(x-7))[/tex]

[tex]8(x -7)^2[/tex]

[tex]c.\ 7a^2x - 6a^2 - 7x + 6[/tex]

Factorize

[tex]a^2(7x - 6) -1( 7x - 6)[/tex]

[tex](a^2 - 1)(7x - 6)[/tex]

The answer can be in this form of further expanded as follows:

[tex](a^2 - 1^2)(7x - 6)[/tex]

Apply difference of two squares

[tex](a+1)(a-1)(7x-6)[/tex]

[tex]d.\ x^4 - x^2 - 12[/tex]

Express [tex]x^4[/tex] as [tex]x^2[/tex]

[tex](x^2)^2 - x^2 - 12[/tex]

Expand

[tex](x^2)^2 +3x^2- 4x^2 - 12[/tex]

[tex]x^2(x^2+3) -4(x^2+3)[/tex]

[tex](x^2-4)(x^2+3)[/tex]

The answer can be in this form of further expanded as follows:

[tex](x^2-2^2)(x^2+3)[/tex]

Apply difference of two squares

[tex](x-2)(x+2)(x^2+3)[/tex]

[tex]e.\ a^{4n} -b^{4n}[/tex]

Represent as squares

[tex](a^{2n})^2 -(b^{2n})^2[/tex]

Apply difference of two squares

[tex](a^{2n} -b^{2n})(a^{2n} +b^{2n})[/tex]

Represent as squares

[tex]((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})[/tex]

Apply difference of two squares

[tex](a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})[/tex]

Answer:

i guess the answer is 680.

Answer:

Step-by-step explanation:

Answer:

mark the answer below as brainliest x

Step-by-step explanation:

Answer:

25 meters per second

Step-by-step explanation:

Divide to find m/s.

750 / 30 = 25

Check in necessary.

25 * 30 = 750

Answer:

25 per sec

Step-by-step explanation:

Answer:

y=(4/3)x-8

Step-by-step explanation:

Slope-intercept form: y=mx+b

4x-3y=24

3y=4x-24 (isolating the 3y)

y=(4/3)x-8 (dividing both sides by 3)

Hope this helps!

Step-by-step explanation:

B. 426 people per square. mile

Answer:

70 I think but I'm not sure

Step-by-step explanation:

AC=70

Answer:

[tex]f(10) = 173[/tex]

Step-by-step explanation:

Given

Exponential Function

[tex](x_1,y_1) = (3,10.125)[/tex]

[tex](x_2,y_2) = (6,34.2)[/tex]

Required

Determine f(10)

We have that

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

First, we need to solve for the values of a and b

For [tex](x_1,y_1) = (3,10.125)[/tex]

[tex]10.125 = ab^3[/tex] --- (1)

For [tex](x_2,y_2) = (6,34.2)[/tex]

[tex]34.2 = ab^6[/tex] ---- (2)

Divide (2) by (1)

[tex]\frac{34.2}{10.125} = \frac{ab^6}{ab^3}[/tex]

[tex]\frac{34.2}{10.125} = \frac{b^6}{b^3}[/tex]

[tex]3.38= b^{6-3}[/tex]

[tex]3.38= b^{3}[/tex]

Take cube root of both sides

[tex]b = \sqrt[3]{3.38}[/tex]

[tex]b = 1.5[/tex]

Substitute 1.5 for b in [tex]10.125 = ab^3[/tex]

[tex]10.125 = a * 1.5^3[/tex]

[tex]10.125 = a * 3.375[/tex]

Solve for a

[tex]a = \frac{10.125}{3.375}[/tex]

[tex]a = 3[/tex]

To solve for f(10).

This implies that x = 10

So, we have:

[tex]y = ab^x[/tex] which becomes

[tex]y = 3 * 1.5^{10[/tex]

[tex]y = 3 * 57.6650390625[/tex]

[tex]y = 172.995117188[/tex]

[tex]y = 173[/tex] -- approximated

Hence:

[tex]f(10) = 173[/tex]

Answer:

2/5

Step-by-step explanation:

(2,2) and (7,4) 4-2/7-2=2/5

Answer (least to greatest):

I hope this helps!

Answer:

1.6,- 11/5 ,2.25 , -2.4 15/10

Step-by-step explanation:

Answer:

-5

Step-by-step explanation:

The equation given is written in slope-intercept form (y = mx + b). In this form, 'm' represents the slope of a line and 'b' represents the y-intercept of the line. Since the equation asks for the slope, we can simply look at the value of 'm' in the equation. Then, we will find the slope.

slope - intercept form

y = mx + b

given equation

-y = -5x + 9

values

m = slope = -5

b = y-intercept = 9

Answer:

The values of given expressions are:

1. gh = -21

2. g^2 - h = 46

3. g + h^2 = 2

4. g + h  = -4

5. h - g  = 10

6. g - h = -10

Step-by-step explanation:

Given values of g and h are:

g = -7

h = 3

1. gh

The two numbers are being multiplied

Putting the values

[tex]gh = (-7)(3) = -21[/tex]

2. g^2-h

Putting the values

[tex]=(-7)^2-3\\=49-3\\=46[/tex]

3. g+h^2

Putting the values

[tex]= -7 + (3)^2\\=-7+9\\=2[/tex]

4. g+h

Putting the values

[tex]= -7+3\\=-4[/tex]

5. h-g

Putting the values

[tex]= 3 - (-7)\\=3+7\\=10[/tex]

6. g-h

Putting values

[tex]=-7-3\\=-10[/tex]

Hence,

The values of given expressions are:

1. gh = -21

2. g^2 - h = 46

3. g + h^2 = 2

4. g + h  = -4

5. h - g  = 10

6. g - h = -10

Answer:

1) - 21

2) 46

3) 2

4) -4

5) 10

6) -10

Step-by-step explanation:

[tex]1)gh = - 7(3) \\ \: \: \: \: \: \: = - 21[/tex]

[tex]2){g}^{2} - h = ( - 7 {)}^{2} - 3 \\ \: \: \: \: \: \: \: \: \: \: \: = 49 - 3 \\ \: \: \: = 46[/tex]

[tex]3)g + {h}^{2} = - 7 + {3}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 7 + 9 \\ \: \: \: \: = 2[/tex]

[tex]4)g + h = - 7 + 3 \\ \: \: \: \: \: \: \: = - 4[/tex]

[tex]5)h - g = 3 - ( - 7) \\ \: \: = 10[/tex]

[tex]6)g - h = - 7 - 3 \\ \: \: \: \: \: \: \: \: \: \: = - 10[/tex]

Answer:

Simplification is a method for solving a system of equations

Step-by-step explanation:

-Graphing is a real way to solve a System of Equations

-Substitution is a real way to Solve Systems of Equations

-Elimination is a real way to solve Systems of Equations

so, Simplifcation can't be one

Take 2x-4y=6 and 4x+2y=17 for example.

They are both in standard form, so you could change them to y=mx+b and graph it. The intersection is the correct answer.

You can also solve one of the equations for y or x, and substitute that into the other equation, and solve for both x and y

or, you could add the two equations together and solve for x and y.

You can't simplify them.

You would multiply each number by 20

Step-by-step explanation:

It says that right there

Answer:

Point C

Happy Holidays!

Step-by-step explanation:

ay

Answer:

(7,4)

Step-by-step explanation:

The x always goes first and the y goes second

Answer:

7,4

Step-by-step explanation:

x,y

Answer:

what is this......................

Answer:

Step-by-step explanation:

Given that:    

The equation of the damped vibrating spring is y" + by' +2y = 0

(a) To convert this 2nd order equation to a system of two first-order equations;

let y₁ = y

y'₁ = y' = y₂

So;

y'₂ = y"₁ = -2y₁ -by₂

Thus; the system of the two first-order equation is:

y₁' = y₂

y₂' = -2y₁ - by₂

(b)

The eigenvalue of the system in terms of b is:

[tex]\left|\begin{array}{cc}- \lambda &1&-2\ & -b- \lambda \end{array}\right|=0[/tex]

[tex]-\lambda(-b - \lambda) + 2 = 0 \ \\ \\\lambda^2 +\lambda b + 2 = 0[/tex]

[tex]\lambda = \dfrac{-b \pm \sqrt{b^2 - 8}}{2}[/tex]

[tex]\lambda_1 = \dfrac{-b + \sqrt{b^2 -8}}{2} ; \ \lambda _2 = \dfrac{-b - \sqrt{b^2 -8}}{2}[/tex]

(c)

Suppose [tex]b > 2\sqrt{2}[/tex], then  λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.

(d)

From λ² + λb + 2 = 0

If b = 3; we get

[tex]\lambda^2 + 3\lambda + 2 = 0 \\ \\ (\lambda + 1) ( \lambda + 2 ) = 0\\ \\ \lambda = -1 \ or \ \lambda = -2 \\ \\[/tex]

Now, the eigenvector relating to λ = -1 be:

[tex]v = \left[\begin{array}{ccc}+1&1\\-2&-2\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right][/tex]

[tex]\sim v = \left[\begin{array}{ccc}1&1\\0&0\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right][/tex]

Let v₂ = 1, v₁ = -1

[tex]v = \left[\begin{array}{c}-1\\1\\\end{array}\right][/tex]

Let Eigenvector relating to  λ = -2 be:

[tex]m = \left[\begin{array}{ccc}2&1\\-2&-1\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right][/tex]

[tex]\sim v = \left[\begin{array}{ccc}2&1\\0&0\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right][/tex]

Let m₂ = 1, m₁ = -1/2

[tex]m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right][/tex]

∴

[tex]\left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= C_1 e^{-t} \left[\begin{array}{c}-1\\1\\\end{array}\right] + C_2e^{-2t} \left[\begin{array}{c}-1/2\\1\\\end{array}\right][/tex]

So as t → ∞

[tex]\mathbf{ \left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= \left[\begin{array}{c}0\\0\\\end{array}\right] \ \ so \ stable \ at \ node \ \infty }[/tex]