Ask question

Ask question

All categories

3 years ago

6

The perimeter of a rectangle is 24 feet. The length is 3 feet less than twice the width. Which of the following systems would be

used to solve this problem ???
The perimeter of a rectangle is 24 feet. The length is 3 feet less than twice the width. The width is:

1 answer:

Temka [501]3 years ago

6 0

the answer would be, "The width is" all of the other information is neccesary!! You could not solve the problem without knowing what the the toal is and what the length is!! You need that info. So "the width is" is not important cuz if you were to take 24 and divide it by 3 * 2 you would have the width.

You might be interested in

the perimeter of a rectangle is 55 inches. the ratio of the length to width is 7:4. use proportions to find the area of the rect

Leni [432]

Answer:

Step-by-step explanation:

let : x the length and y the width .... ( x > y and x ; y reals numbers )

the perimeter is : 2(x+y)

2(x+y) = 55...(1)

x/y = 7/4 ...(2)

by (2) : x = (7/4)y

subsct in (1) : 2( (7/4)y +y ) =55

(7/2)y +2y =55

multiply by 2

7y +4y = 110

11y = 110

y = 10

x = (7/4)(10)

x = 35/2

4 0

3 years ago

Which expressions are equivalent to 4x^4-9x^2?

weeeeeb [17]

We can cosider this to be a difference of 2 squares so

4x^4 - 9x^2 = (2x^2 - 3x)(2x^2 + 3x) so D is one answer

also we could take x^2 out and get
x^2(4x^2 - 9) = C

3 0

3 years ago

Write each equation in slope intercept form 9x + 35 = -5y. 2y - 6 = -6x. -11 - 7y = -56

tamaranim1 [39]

Answer:

y= -9/5x -7

y= -3x +2

y= 45/7

6 0

2 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Mathematics and fractions 1/2 and 6/12

77julia77 [94]

Answer:

Equivalent fractions

Step-by-step explanation:

7 0

3 years ago