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3 years ago

Area is 20 square units. the length is 4 more than 3 time the width. find the dimension

2 answers:

8 0

starts with 20 smthing x3 +4 = 20 but what is it ???? hmmmm if you think more you can fiqure it out by just finding random numbers so use ur brain i guess???

dezoksy [38]3 years ago

4 0

Answer:

width: 2

length: 10

Step-by-step explanation:

L = 3W + 4

A = LW = (3W + 4) x W

= 3w² +4w

3w² + 4w = 20

3w² + 4w -20 = 0

(w - 2) (3w + 10) = 0

w = 2 or w = - 10/3 (width is positive) ∴ width = 2

length = 3 x 2 + 4 = 10

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Determine whether the relation describe c as a function of w.

Delicious77 [7]

Answer:

Function:

c = f(w) = 0.49, 0 < w ≤ 1

= 0.70, 1 < w ≤ 2

= 0.91, 2 < w ≤ 3

Step-by-step explanation:

Yes, the relation described can be interpreted as a function.

Here, c is the cost of a mail letter. c depends upon w, which is the weights of the mail letter.

As described in the question, the relation can be expressed as a function.

c can be expressed as a function of w in the following manner:

c(cost of mail) = f(w), where w is the independent variable and c is the dependent variable

c = f(w) = 0.49, 0 < w ≤ 1

= 0.70, 1 < w ≤ 2

= 0.91, 2 < w ≤ 3

where, c is in dollars and w is in ounces.

5 0

3 years ago

Use the graph to answer the questions help!

max2010maxim [7]

Answer:

0; 2; 6; 0 or -1

Step-by-step explanation:

If the question says "f(a)=" they are asking for the y value with the given x such as the two first ones.

If it's "f(x)=a than x=" they are asking you what x is when the (bolded) x is a given value.

5 0

3 years ago

-8a-9a+7a what does this equal

allochka39001 [22]

The answer would be -10a

7 0

3 years ago

Sammy can run 3 1/2 feet in 2 seconds what is his unit rate ( feet per second)

zlopas [31]

1.75ft/second. 3.5/2 simplifies to 1.75/1.

7 0

3 years ago

Read 2 more answers

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

5 0

3 years ago