Math algebra 2 show you’re work plz

Answer:

34m = c

Step-by-step explanation:

For every month (m) you pay 34 dollars. However many months youu use that service time 34 equals your total cost (c).

Answer:

[tex]let \: cost \: be \: { \bf{c}} \: and \: months \: be \: { \bf{n}} \\ { \bf{c \: \alpha \: n}} \\ { \bf{c = kn}} \\ 34 = (k \times 1) \\ k = 34 \: dollars \\ \\ { \boxed{ \bf{c = 34n}}}[/tex]

Answer:

Growth: y = 1700(1.25)^t, y = 4000(1.0825)^t, y = 1.5(10)^t

Decay: y =240(1/2)^t, y = 12,000(0.72)^t,  y = 8000(0.97)^t

Step-by-step explanation:

In an exponential equation, growth and decay are determined by the factor you are multiplying by exponentially. If it's under 1 you're basically exponentially dividing the initial value. Over 1 and you are increasing the value.

the graph would steeper, meaning more savings over time

Answer:

The solution to the equation  log3(x+2)=1 is given by x=1

Step-by-step explanation:

We are given that

[tex]log_3(x+2)=1[/tex]

We have to find the graph which shows the  solution to the equation log3(x+2)=1.

[tex]log_3(x+2)=1[/tex]

[tex]x+2=3^1[/tex]

Using the formula

[tex]lnx=y\implies x=e^y[/tex]

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

[tex]x=3-2[/tex]

[tex]x=1[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

The graph shows the function value is zero for x=5 and x=7. These are the elements of the solution set.

  x ∈ {5, 7}

__

The graph is below the x-axis between these points, so that is the region where f(x) < 0

  5 < x < 7 . . . . . for f(x) < 0

In interval notation: (5, 7).

Answer:

It has a difference of x=2 of -4

Step-by-step explanation:

It has a difference of x=2 of -4

Given ,

           x ≠ 2 ,

           x - 2 =0

So, we put x = -2 because in question x ≠ 2 .

Then,      x - 2 = 0

              -2 -2 = 0

               - 4 =0

Therefore, it has a difference of x= -2 of -4.

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Answer:

[tex]9 + 7i[/tex]

Step-by-step explanation:

[tex]7i^5+9i^8[/tex]

[tex]i^5 = i\\ i^8 = 1[/tex]

Answer:

6

Step-by-step explanation:

We have the expression "3x is 18" and we want to find out x, so the easiest way is to isolate x. The way to do this is by dividing both 3x and 18 by 3, leaving us with 6. This means that our answer for x is 6. I hope this helped and please don't hesitate to reach out for clarification!

Answer:

6

Step-by-step explanation:

3x = 18

Divide both sides by 3

3x / 3 ( the 3s cancel out and we're left with x )

18/3 = 6

We get that x = 6

We can then plug in 6 to x in 3x = 18 to check if our answer is correct

3x = 18

Replace x with 6

3(6) = 18

3(6) = 18

18 = 18

This is true hence we can confirm that x = 6

Answer:

The area of the square is increasing at a rate of 40 square centimeters per second.

Step-by-step explanation:

The area of the square ([tex]A[/tex]), in square centimeters, is represented by the following function:

[tex]A = l^{2}[/tex] (1)

Where [tex]l[/tex] is the side length, in centimeters.

Then, we derive (1) in time to calculate the rate of change of the area of the square ([tex]\frac{dA}{dt}[/tex]), in square centimeters per second:

[tex]\frac{dA}{dt} = 2\cdot l \cdot \frac{dl}{dt}[/tex]

[tex]\frac{dA}{dt} = 2\cdot \sqrt{A}\cdot \frac{dl}{dt}[/tex] (2)

Where [tex]\frac{dl}{dt}[/tex] is the rate of change of the side length, in centimeters per second.

If we know that [tex]A = 25\,cm^{2}[/tex] and [tex]\frac{dl}{dt} = 4\,\frac{cm}{s}[/tex], then the rate of change of the area of the square is:

[tex]\frac{dA}{dt} = 2\cdot \sqrt{25\,cm^{2}}\cdot \left(4\,\frac{cm}{s} \right)[/tex]

[tex]\frac{dA}{dt} = 40\,\frac{cm^{2}}{s}[/tex]

The area of the square is increasing at a rate of 40 square centimeters per second.

answer choice a

x^2 - 5x - 1 = 0

a=1, b=-5, and c=-1

next you plug in the numbers into the quadratic formula

-(-5) plus or minus the square root of (-5)^2 - 4(1)(-1)/ 2(1)

after you simply, you should get 5 plus or minus the square root of 29, which is answer choice a

Answer:

= ∑ 6*n*x^n-1

Radius of convergence = 1

Step-by-step explanation:

f(x) = 6(1-x)^-2

Maclaurin series can be expressed using the formula

f(x) =  f(0) + f '(0)x + f ''(0)/ 2!  (x)^2 + f '''(0)/3! (x)^3 + f (4)(0) 4! x4 + .

attached below is the detailed solution

Radius of convergence = 1

The Maclaurin series for f(x) = 6 / (1 - x )^2  = ∑ 6*n*x^n-1  ( boundary ; ∞ and n = 1 )

Answer:

90 cause 90 times 10 is 900 and 10% times 10 is 100%

Step-by-step explanation:

Answer: 90

Step-by-step explanation: 900 × .10 = 90

Answer:

30 gallons of each brand

Step-by-step explanation:

1 gallon of 60% + 1 gallon of 80% = 2 gallons 70% (60/2 = 30)

Answer:

The first one is 2

The second one is 0

I hope this helps!

Answer:

First:

[tex]{ \tt{ = 4 + ( - 2)}} \\ = 4 - 2 \\ = 2[/tex]

Second:

[tex] - 3 + 3 \\ = 0[/tex]

Answer:

C

Step-by-step explanation:

The triangles are similar and the scale factor can be find out by computing 18/6=15/5=9/3=3

Answer:

[tex]\displaystyle V = \frac{625 \pi}{2}[/tex]

General Formulas and Concepts:

Algebra I

Calculus

Integrals

Integration Rule [Reverse Power Rule]:                                                               [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:                                     [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Shell Method:                                                                                                         [tex]\displaystyle V = 2\pi \int\limits^b_a {xf(x)} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

y = x²

y = 0

x = 5

Step 2: Identify

Find other information from graph.

See Attachment.

Bounds of Integration: [0, 5]

Step 3: Find Volume

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Applications of Integration

Book: College Calculus 10e

Answer:

[tex]\cos(\theta)= \frac{\sqrt{377}}{29}[/tex]

Step-by-step explanation:

Given

[tex]\sin(\theta) = -\frac{4}{\sqrt{29}}[/tex] -- the correct expression

Required

[tex]\cos(\theta)[/tex]

We know that:

[tex]\sin^2(\theta) + \cos^2(\theta)= 1[/tex]

Make [tex]\cos^2(\theta)[/tex] the subject

[tex]\cos^2(\theta)= 1 - \sin^2(\theta)[/tex]

Substitute: [tex]\sin(\theta) = -\frac{4}{\sqrt{29}}[/tex]

[tex]\cos^2(\theta)= 1 - (-\frac{4}{\sqrt{29}})^2[/tex]

Evaluate all squares

[tex]\cos^2(\theta)= 1 - (\frac{16}{29})[/tex]

Take LCM

[tex]\cos^2(\theta)= \frac{29 - 16}{29}[/tex]

[tex]\cos^2(\theta)= \frac{13}{29}[/tex]

Take square roots of both sides

[tex]\cos(\theta)= \±\sqrt{\frac{13}{29}}[/tex]

cosine is positive in the 4th quadrant;

So:

[tex]\cos(\theta)= \sqrt{\frac{13}{29}}[/tex]

Split

[tex]\cos(\theta)= \frac{\sqrt{13}}{\sqrt{29}}[/tex]

Rationalize

[tex]\cos(\theta)= \frac{\sqrt{13}}{\sqrt{29}} * \frac{\sqrt{29}}{\sqrt{29}}[/tex]

[tex]\cos(\theta)= \frac{\sqrt{13*29}}{29}[/tex]

[tex]\cos(\theta)= \frac{\sqrt{377}}{29}[/tex]

9514 1404 393

Answer:

  x² +y² -6x -16y +48 = 0

Step-by-step explanation:

Given:

Find:

  general form equation for the circle

Solution;

The standard form equation for the circle is ...

  (x -h)² +(y -k)² = r² . . . . . circle with radius r centered at (h, k)

  (x -3)² +(y -8)² = 5²

Subtracting 25 and expanding this will give the general form.

  x² -6x +9 +y² -16y +64 -25 = 0

  x² +y² -6x -16y +48 = 0

_____

Additional comment

"General form" of an equation is usually the form f(x,y) = 0, where f(x, y) is written in "standard form," with terms in lexicographical order and decreasing degree.

Answer:

-5 it is

Step-by-step explanation:

Given:

The graph of a function.

To find:

The range of the given function using interval notation.

Solution:

Range: The set of y-values or output values are known as range.

From the given graph, it is clear that the function is defined for [tex]0<x<4[/tex] and the values of the functions lie between -2 and 2, where -2 is excluded and 2 is included.

Range [tex]=\{y|-2<y\leq 2\}[/tex]

The interval notation is:

Range [tex]=(-2,2][/tex]

Therefore, the range of the given function is (-2,2].

Answer:

[tex]90[/tex] [tex]ft^3[/tex]

Step-by-step explanation:

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

The formula to find the volume of a rectangular prism is   [tex]V=lwh[/tex]

Let's substitute the number for the length, width, and height now.

[tex]V=(6)(5)(3)[/tex]

[tex]V=(30)(3)[/tex]

[tex]V=90[/tex]

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

Hope this is helpful.

Answer:

(-2,3,10,8)

Step-by-step explanation:

eg

(x,y)=(domain,range)

x components are domain and y components are range in the given set of ordered pairs

Answer:

144

Step-by-step explanation:

We can write this as 2 shirts = 16.00 . Then, we can write this as a ratio, where

2 shirts / 16.00 = 18 shirts/ cost of 18 shirts as the cost for a pair of shirts remains the same, so the ratio will as well.

Therefore, writing the cost of 18 shirts as c, we have

2 shirts/16.00 = 18 shirts/c

multiply both sides by c to remove a denominator

2 shirts * c/16.00 = 18 shirts

multiply both sides by 16 to remove the other denominator

2 shirts * c = 18 shirts * 16

divide both sides by 2 shirts to isolate the c, as we want to figure that out

c = 18 shirts * 16/2 shirts

c=144

admins, pls delete this, I messed up and don't know how

Answer:

the answer should be e^3

Step-by-step explanation:

i hope it helps you

Answer: The ans is imaginary . option c

Answer:

Step-by-step explanation:

a = 5

b = 15

c = ?

c^2 = a^2 + b^2

c^2 = 5^2 + 15^2

c^2 = 25 + 125

c^2 = 250

sqrt(c^2) = sqrt(250)

c = sqrt(250)

Answer: hi your question is poorly written below is the correct question

answer :

a) y1 = Asint,   y'1 = Acost  , y"1 = -Asint

b) y2 = Bcost,   y'2 = Bsint , y"2 = - Bcost

c) y = Asint + B cost satisfies the differential equation for any constant A and B

Step-by-step explanation:

y" + y = 0

Proves

a) y1 = Asint,   y'1 = Acost  , y"1 = -Asint

b) y2 = Bcost,   y'2 = Bsint , y"2 = - Bcost

c) y3 = y1 + y2 ,   y'3 = y'1 + y'2,  y"3 = y"1 + y"2

∴ y"1 + y1 = -Asint + Asint

  y"2 + y2 = -Bcost + Bcost

  y"3 - y3 = y"1 + y"2 - ( y1 + y2 )

               = y"1 - y1 + y"2 - y2  

               = -Asint - Asint  + ( - Bcost - Bcost )  = 0

Hence we can conclude that y = Asint + B cost satisfies the equation for any constant A and B