The perimeter of a rectangle is 13cm and its width is 11/4. find its length.

1 answer:

6 0

Perimeter = 2 length + 2 width

P = 2(l) + 2(11/4)

$13 = 2l +2* \frac{11}{4}$

$13 = 2l + \frac{2}{1} * \frac{11}{4}$

$13 = 2l + \frac{22}{4}$ (subtract 22/4 from each side)

$13 -\frac{22}{4} = 2l$

$13 -\frac{11}{2} = 2l$

$\frac{26}{2} -\frac{11}{2} = 2l$

$\frac{15}{2}} = 2l$

$7.5 = 2l$ (divide each side by 2)

7.5 ÷ 2 = l

3.75 = l

The answer is 3.75 or $3 \frac{3}{4}$

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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$