The value of a television t years
after its purchase is modeled by the
exponential function ƒ(t) =
1500(0.83)t. Which best describes
the domain and range of this
exponential function?

Answer:

y = -3x - 1

hope this helps!

Answer:

Step-by-step explanation:

Answer:

[tex]32\sqrt{7}-24\sqrt{2}[/tex]

Step-by-step explanation:

[tex]8(4\sqrt{7}-\sqrt{18})\\\\8(4\sqrt{7}-3\sqrt{2})\\\\32\sqrt{7}-24\sqrt{2}[/tex]

[tex]\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}[/tex]

Given equation -

[tex]\bold{(x + 2) {}^{2} + 4(x + 2) - 5 = 0} \\ [/tex]

solve the parenthesis so as to obtain simpler terms

[tex]\bold{\implies \: ( x {}^{2} + 4x + 4 ) + 4x + 8 - 5 = 0}\\ [/tex]

solve the like terms and you'll obtain the required equation !

[tex]\bold{\implies \: x {}^{2} + 8x - 7 = 0}[/tex]

hope helpful ~

[tex]{\large{\red{\mapsto{\maltese{\underline{\green{\boxed{\blue{\underbrace{\overbrace{\pink{\pmb{\bf{Answer:}}}}}}}}}}}}}}[/tex]

x = -7, -1

Step-by-step explanation:

[tex] \sf {(x + 2)}^{2} + 4(x + 2) - 5 = 0[/tex]

let (x + 2) = y

So equation will be

[tex] \sf {y}^{2} + 4y - 5 = 0[/tex]

Further it becomes quadratic equation

we will solve it by breaking middle term method

[tex] \sf \implies {y}^{2} + 5y - y - 5 = 0 \\ \\ \sf \implies y(y + 5) - 1(y + 5) = 0 \\ \\ \sf \implies (y + 5)(y - 1) = 0[/tex]

Now put the value of y

[tex] \sf \implies (x + 2 + 5)(x + 2 - 1) = 0 \\ \\ \sf \implies (x + 7)(x + 1) = 0[/tex]

(x + 7) =0⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(x + 1) = 0

x = -7⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ x = -1

Answer: D. Using another person's ideas or writing without citing them.

Step-by-step explanation: Plagiarism is the use of someones work and claiming it as their own without giving them credit

A Line graph

Sure hope this helps you

Answer:

Option C is the correct answer

Step-by-step explanation:

The sum of the expression (x - 2) and (x+6) is 2x + 8

The sum of numbers simply means to add up the numbers.

Therefore,

(x  -2) + (x + 6)

Hence,

(x - 2) + (x + 6)

open the brackets

Therefore,

x - 2 + x + 6

combine like terms

x + x + 2 + 6

add or subtract the like terms

2x + 8

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Answer:

Area = 50.125 cm^2

Perimeter = 26 + 15 = 41 cm

Step-by-step explanation:

Area Rectangle = 6 * 7 = 42 cm^2

Area Triangle = 1/2 * (9.5 - 7) * 6.5

Area Triangle = 1/2 * (2.5) * 6.5 = 8.125 cm^2

Total Area = 42 + 8.125 = 50.125 cm^2

Perimeter Rectangle = 6 + 7 + 6 + 7 = 26 cm

Perimeter or Triangle = 2.5 + 6.5 + 6 = 15 cm

Total Perimeter = 26 + 15 = 41 cm

When you use word processor functions such as spell check and thesaurus to write a second draft, you are "C. Revising."

Revising means making changes and improvements to a piece of writing to make it better. This includes making it easier to understand, organizing it well, and making sure it is good quality overall.

Using spell check helps you fix mistakes in your spelling, while using the thesaurus helps you find different words that mean the same thing and choose better words to make your writing clearer and more accurate.

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Answer:

$1,129.27

Step-by-step explanation:

Compounded interest formula is

[tex]A=P(1+\frac{r}{n} )^{nt}[/tex]

Where [tex]A[/tex] is the final amount, [tex]P[/tex] is the principal, [tex]r[/tex] is the anual interest in decimal, [tex]n[/tex] is the numer of compounded periods in one year and [tex]t[/tex] is the time in years.

[tex]P= 800\\r= 0.09\\n= 1\\t= 4[/tex]

So then we plug our numbers...

[tex]A=800(1+\frac{0.09}{1})^{1(4)}=800(1.09)^{4}[/tex]

[tex]A=$1,129.27[/tex]

After 4 years the amount would be $1,129.27

Answer:

milk because ur dad left the milk

Step-by-step explanation:

ur mum bc she is a pro statue

Answer:

A lot

Step-by-step explanation:

So basically you learn a lot

Answer:

Round off value = 500000

Step-by-step explanation:

467532  > 500

- Round up

⇒ 500,000

Answer:

D.72

Step-by-step explanation:

first you have to divide 96 and 4 than it will equal 24 than you multiple by 3 and so 24×3=72

Answer:

a) x=loge3 ( log e base 3)

b)4=e^y

Answer:

A.) loge(x)=3

B.)  e^y =4

Step-by-step explanation:

A.) loge(x)=3

B.)  e^y =4

The Cost of Carl's first item is 3.95. The Cost of his second item is 4.95. To find the total amount we need to add these two numbers together.

3 plus 4 is 7

95 cents plus 95 cents is 1.90

7+1.90 is 8.90

The total cost is $8.90

Hope this helps!

Answer:

3.5 centimeters

Step-by-step explanation:

75% of 2 cm = .75 x 2 = 1.5 cm

2 cm + 1.5 cm = 3.5 cm

Answer:

Answer with explanation is below~

Step-by-step explanation:

We could easily obtain the formulae/formula of an equilateral triangle not only through Pythagoras Theorem, but also through Heron's formula.

INTRODUCTION:

What's equilateral triangle?

What's Heron Formula?

[tex] \boxed{\rm \: Area= \sqrt{s(s - a)(s - b)(s - c)} \: units ^{2}} [/tex]

Where,

What's Pythagoras' Theorem?

SOLVING:

Let,ABC an equilateral triangle of sides a.

Now:Draw a perpendicular straight line AM to the side BC(Name each part of triangle)

So it's clear that ∆AMB is a right angled triangle at M, BM = (1/2)BC = a/2.

Please note AM here represents the height of ∆ ABC.

[tex] \boxed{\rm \: AM = \sqrt{AB^2-BM^2}}[/tex]

[tex] \rm \: AM = \sqrt{a {}^{2} - { \bigg( \cfrac{a}{2} \bigg) }^{2} } [/tex]

[tex] \rm \: AM = 3 \: \cfrac{a {}^{2} }{4} [/tex]

[tex] \rm \: AM = \cfrac{ \sqrt{3} }{2} \: a[/tex]

Now find the area of ∆ABC:

[tex] \rm \triangle \: ABC = \cfrac{1}{2} \times \: BC \times A [/tex]

[tex] \rm \: \triangle \: ABC = \cfrac{1}{2} \times a \times \cfrac{ \sqrt{3}}{4}a [/tex]

[tex] \boxed{\rm\triangle \: ABC = \cfrac{ \sqrt{3 } }{4} a {}^{2} \: units {}^{2}} [/tex]

Hence,the formulae of equilateral triangle using Pythagoras' theorem is {√(3)/4} a^2

[tex] \rule{225pt}{2pt}[/tex]

Extras:

Now let's find the area using Heron's Formula.

Solving:

Let each side of an equilateral triangle be a.

SO, then:

s = (3a/2)

We know that, (Heron's formula)

[tex] \rm \: A = \sqrt{s(s - a)(s - b)(s - c)} [/tex]

Now the area A :

[tex]\rm AR(A)= \sqrt{ \cfrac{ 3a } 2 \bigg(\cfrac{ 3a }{2} - a \bigg)\bigg(\cfrac{ 3a }{2} - a \bigg)\bigg(\cfrac{ 3a }{2} - a \bigg)} [/tex]

[tex]\rm AR(A)= \sqrt{ \cfrac{3a}{2} \bigg( \cfrac{a}{2} \bigg)\bigg( \cfrac{a}{2} \bigg)\bigg( \cfrac{a}{2} \bigg)}[/tex]

[tex]\rm \boxed{ \rm AR(A)= \cfrac{ \sqrt{ 3a}}{4} \: a {}^{2} \: \: units {}^{2}} [/tex]

And voila! we're done!

I hope this helps! :)

[Figure of Equilateral triangle is attached, it denotes the triangle we need to draw while finding the formulae through Pythagoras' theorem. ]

44 rem

Im sorry if its wrong

Answer:

[tex]\displaystyle \boxed{ \frac{dz}{du} = \frac{e^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 4u + v \Bigg] }[/tex]

[tex]\displaystyle \boxed{ \frac{dz}{dv} = \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 1 - \frac{2u + v}{v} \Bigg] }[/tex]

General Formulas and Concepts:

Calculus

Differentiation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

Multivariable Calculus

Partial Derivatives

Partial Derivative Rule [Chain Rule]:
[tex]\displaystyle \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial t}[/tex]

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle x = 2u + v[/tex]

[tex]\displaystyle y = \frac{u}{v}[/tex]

[tex]\displaystyle z = e^{xy}[/tex]

Step 2: Find Derivatives

[tex]\displaystyle \frac{dz}{du}[/tex]:

We can find the partial derivatives by differentiating using basic differentiation rules found under "Calculus":

[tex]\displaystyle\begin{aligned}\frac{\partial z}{\partial x} & = ye^{xy} \\\frac{\partial x}{\partial u} & = 2 \\\frac{\partial z}{\partial y} & = xe^{xy} \\\frac{\partial y}{\partial u} & = \frac{1}{v} \\\end{aligned}[/tex]

Substituting in our partial derivatives and our variables x and y, we can obtain a final derivative:

[tex]\displaystyle\begin{aligned}\frac{dz}{du} & = 2ye^{xy} + \frac{xe^{xy}}{v} \\& = \frac{2ue^\big{\frac{u(2u + v)}{v}}}{v} + \frac{(2u + v)e^\big{\frac{u(2u + v)}{v}}}{v} \\& = \boxed{ \frac{e^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 4u + v \Bigg] }\end{aligned}[/tex]

∴ we have found the derivative of z with respect to u.

[tex]\displaystyle \frac{dz}{dv}[/tex]:

We can find the partial derivatives by using the same method(s):

[tex]\displaystyle\begin{aligned}\frac{\partial z}{\partial x} & = ye^{xy} \\\frac{\partial x}{\partial v} & = 1 \\ \frac{\partial z}{\partial y} & = xe^{xy} \\\frac{\partial x}{\partial v} & = \frac{-u}{v^2} \\ \end{aligned}[/tex]

Substituting in our partial derivatives and our variables x and y, we can obtain a final derivative:

[tex]\displaystyle\begin{aligned}\frac{dz}{dv} & = ye^{xy} + \frac{-uxe^{xy}}{v^2} \\& = \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} + \frac{-u(2u + v)e^\big{\frac{u(2u + v)}{v}}}{v^2} \\& = \boxed{ \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 1 - \frac{2u + v}{v} \Bigg] }\end{aligned}[/tex]

∴ we have found the derivative of z with respect to v.

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Topic: Multivariable Calculus

Unit: Partial Derivatives and Applications

Answer:

2:3

Step-by-step explanation:

Simplica Los dos numeros

6 ÷ 3 = 2

9 ÷ 3 = 3

Answer:

  slope 4/5

Step-by-step explanation:

The attachment shows the rise in green (4 grid squares), and the run in red (5 grid squares). The slope is the ratio rise/run.

__

The slope is rise/run = 4/5.

Using proportions, it is found that the dimensions of the reception area of of 12 feet by 20 feet.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

In this problem, the scale factor is of 1 inch = 4 feet, meaning that each inch on the drawing represents a dimension of 4 feet.

The drawing is of 3 inches by 5 inches, hence the real dimensions are given by;

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Answer:

See below

Step-by-step explanation:

Not sure what they are looking for here....unless there is more info

[tex]\sqrt[14]{y}[/tex]  =  [tex]y^{1/14}[/tex]     perhaps?

Answer:

2

Step-by-step explanation:

first, add terms

8 + 5 + 12 + 4 + 5 + 8 + 7 = 49

seven terms, therefore we divide by 7

49/7 = 7

find distance

8-7= 1

7-5= 2

12-7=5

7-4=3

7-5=2

8-7=1

7-7=0

added together you get 14,

14/7 = 2

your answer is 2

hope this helps:)

A mean is an arithmetic average of a set of observations. The height of the 9th-grade students will be 66.7±4 inches.

A mean is an arithmetic average of a set of observations. it is given by the formula,

Mean = (Sum of observations)/Number of observations

Given the height of the 10 selected students is {66, 61, 67, 61, 75, 71, 70, 61, 61, and 74}, also, the standard deviation of the 9th-grade students is 4 inches. Therefore, The mean height in inches of the selected students is,

Mean = (66+61+67+61+75+71+70+61+64)/10 = 66.7inches

Therefore, the height of the 9th-grade students will be 66.7±4 inches.

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Answer:

BC = 6 units

Given following:

Solve for BC:

Formula:

[tex]\rightarrow \sf tan(A) =\dfrac{opposite}{adjacent}[/tex]

insert values and make them proportional

[tex]\rightarrow \sf tan(A) =\dfrac{BC}{AC} = \dfrac{3}{2} = \dfrac{6}{4}[/tex]

So, BC is 6 units.

Answer:

1)  BC = 6

2)  B = 33.7°  (nearest tenth)

3)  [tex]\sf AB=2\sqrt{13}[/tex]

Step-by-step explanation:

Trigonometric ratios

[tex]\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}[/tex]

where:

Given:

Substituting the values into the tan ratio:

[tex]\implies \sf \tan A=\dfrac{BC}{AC}=\dfrac{BC}{4}[/tex]

Given:

[tex]\sf \tan A=\dfrac{3}{2}[/tex]

Therefore:

[tex]\implies \tan \sf A = \tan \sf A[/tex]

[tex]\implies \sf \dfrac{BC}{4}=\dfrac{3}{2}[/tex]

[tex]\implies \sf BC=\dfrac{4 \cdot 3}{2}=6[/tex]

Given:

Substituting the values into the tan ratio and solving for B:

[tex]\implies \sf \tan B=\dfrac{AC}{BC}[/tex]

[tex]\implies \sf \tan B=\dfrac{4}{6}[/tex]

[tex]\implies \sf B=\tan ^{-1}\left(\dfrac{4}{6}\right)[/tex]

[tex]\implies \sf B=33.69006753...^{\circ}[/tex]

[tex]\implies \sf B=33.7^{\circ}\:(nearest\:tenth)[/tex]

Pythagoras’ Theorem

[tex]a^2+b^2=c^2[/tex]

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

Substituting the values into the formula and solving for AB:

[tex]\implies \sf AC^2+BC^2=AB^2[/tex]

[tex]\implies \sf 4^2+6^2=AB^2[/tex]

[tex]\implies \sf 16+36=AB^2[/tex]

[tex]\implies \sf AB^2=52[/tex]

[tex]\implies \sf AB=\sqrt{52}[/tex]

[tex]\implies \sf AB=2\sqrt{13}[/tex]