If f(x)=8x then what is the area enclosed by the graph of the function, the horizontal axis, and vertical lines at x=2 and x=6

Answers

Answer 1

Answer:

Answer:

$A=128$

Step-by-step explanation:

First of all we need to graph f(x)=8x, (First picture)

Now we have to calculate the area enclosed by the graph of the function, the horizontal axis, and vertical lines at $x_(1)=2$ and $x_(2)=6$ ,

The area that we have to calculate is in pink (second picture).

The function is positive and the domain is $[2,6]$ then we can calculate the area with this formula:

$A=\int\limits^b_a {f(x)} \, dx$ ,

In this case $b=x_(2) , a=x_(1)$

$A=\int\limits^6_2 {8x} \, dx = 8\int\limits^6_2 {x} \, dx$

The result of the integral is,

$A=8(x^(2))/(2)$ , but the integral is defined in [2,6] so we have to apply Barrow's rule,

Barrow's rule:

If f is continuous in [a,b] and F is a primitive of f in [a,b], then:

$\int\limits^b_a {f(x)} \, dx =F(b)-F(a)$

Applying Barrow's rule the result is:

$A=8.(6^(2) )/(2)-8.(2^(2) )/(2)$

$A=8.(36)/(2) -8.(4)/(2)$

$A=144-16$

$A=128$

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a recipe for salsa is 65% tomatoes.if a batch contains 5.2 cups of tomatoes,how large is the batch of salsa. please help!!

Answers

Answer:

8 cups

Step-by-step explanation:

Let the total batch of salsa be x.

$(65)/(100)$ x = 5.2

$(100)/(100)$ x = $(100)/(65)$ × 5.2

x = 8

8 cups of tomatoes...

H3lp pl34ssssssssssssssssssssssssss

Answers

Answer:

Step-by-step explanation:

$(a^(2)-1)/(2-5a) times (15a-6)/(a^(2)+5a-6)$ click on answer to see full problem

Answers

$\bf \cfrac{a^2-1}{2-5a}* \cfrac{15-6}{a^2+5a-6}\n\n-----------------------------\n\nrecall\quad \textit{difference of squares}\n \quad \n(a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\n\nthus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)\n\n\nnow\quad a^2+5a-6\implies (a+6)(a-1)\n\n-----------------------------\n\nthus\n\n\n\cfrac{a^2-1}{2-5a}* \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}* \cfrac{3(5a-2)}{(a+6)(a-1)}\n\n-----------------------------\n\n$

$\bf now\quad 3(5a-2) \iff -3(2-5a)\n\n-----------------------------\n\nthus\n\n\n\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}* \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}$

HELP PLZ ASAP!!!!A landfill has 60,000 tons of waste in it. Each month it accumulates an average of 360 more tons of waste. What is a function rule that represents the total amount of waste after m months? Let W represent the total amount of waste and m represent months.

A. W= 360m + 60,000

B. W= 60,000m + 360

C. W= 360m - 60,000

D. W= 60,000m - 360

Answers

Answer:

Step-by-step explanation:

Your answer is A.

Anytime you have a variable or something that changes such as the month that is a variable so that means its going to be your x/m in this case

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers y of cell sites from 1985 through 2011 can be modeled byy = 269573/1+985e^-0.308t where t represents the year, with t = 5 corresponding to 1985. Use the model to find the numbers of cell sites in the years 1998, 2003, and 2006.

Answers

Answer:

(a) 3178

(b) 14231

Step-by-step explanation:

Given

$y = (269573)/(1+985e^(-0.308t))$

Solving (a): Year = 1998

1998 means t = 8 i.e. 1998 - 1990

So:

$y = (269573)/(1+985e^(-0.308*8))$

$y = (269573)/(1+985e^(-2.464))$

$y = (269573)/(1+985*0.08509)$

$y = (269573)/(84.81365)$

$y = 3178$ --- approximated

Solving (b): Year = 2003

2003 means t = 13 i.e. 2003 - 1990

So:

$y = (269573)/(1+985e^(-0.308*13))$

$y = (269573)/(1+985e^(-4.004))$

$y = (269573)/(1+985*0.01824)$

$y = (269573)/(18.9664)$

$y = 14213$ --- approximated

Solving (c): Year = 2006

2006 means t = 16 i.e. 2006 - 1990

So:

$y = (269573)/(1+985e^(-0.308*16))$

$y = (269573)/(1+985e^(-4.928))$

$y = (269573)/(1+985*0.00724)$

$y = (269573)/(8.1314)$

$y = 33152$ --- approximated

What is the slope of the function? –6 –4 4 6graph :
x
-2
-1
0
1
2

y
8
2
-4
-10
-16

Answers

to get the slope, all we need is two points, so let's pick two off the table.

$\bf \setlength{\fboxsep}{1pt}\begin{array}{rrll}x&y\n\cline{1-2}\n\boxed{-2}&\boxed{8}\n-1&2\n\boxed{0}&\boxed{-4}\n1&-10\n2&-16\end{array}~\hspace{5em}(\stackrel{x_1}{-2}~,~\stackrel{y_1}{8})\qquad(\stackrel{x_2}{0}~,~\stackrel{y_2}{-4})\n\n\n\slope = m\implies\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-8}{0-(-2)}\implies \cfrac{-4-8}{0+2}\implies \cfrac{-12}{2}\implies -6$

Answer:

-6 i got it right

Step-by-step explanation:

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