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3 years ago

Need help with area for solving this

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

8 0

Answer:

Step-by-step explanation:

First find the area of total rectangle which equals 20 * 12 = 240

Second find the area of triangle which equals 0.5 * 4 * 6 = 12

The required area = total rectangle - triangle = 240 - 12 = 228

Bumek [7]3 years ago

8 0

Answer:

NEEEEEGARRRRR

Step-by-step explanation:

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A cardboard box without a lid is to have a volume of 19,652 cm3. Find the dimensions that minimize the amount of cardboard used.

Effectus [21]

Answer:

The dimension of the cardboard is 34 cm by 34 cm by 17 cm.

Step-by-step explanation:

Let the dimension of the cardboard box be x cm by y cm by z cm.

The surface area of the cardboard box without lid is

f(x,y,z)= xy+2xz+2yz.....(1)

Given that the volume of the cardboard is 19,652 cm³.

Therefore xyz =19,652

$\Rightarrow z=\frac{19652}{xy}$ ......(2)

putting the value of z in the equation (1)

$f(x,y)=xy+2x(\frac{19652}{xy})+2y(\frac{19652}{xy})$

$\Rightarrow f(x,y)=xy+\frac{39304}{y})+\frac{39304}{x}$

The partial derivatives are

$f_x=y-\frac{39304}{x^2}$

$f_y=x-\frac{39304}{y^2}$

To find the dimension of the box set the partial derivatives $f_x=0$ and $f_y=0$ .Therefore $y-\frac{39304}{x^2}=0$

$\Rightarrow y=\frac{39304}{x^2}$ .......(3)

and $x-\frac{39304}{y^2}=0$

$\Rightarrow x=\frac{39304}{y^2}$ .......(4)

Now putting the x in equation (3)

$y =\frac {39304}{(\frac{39304}{y^2})^2}$

$\Rightarrow y=\frac{y^4}{39304}$

$\Rightarrow y^3= 39304$

⇒y=34 cm

Then $\Rightarrow x=\frac{39304}{34^2}$ =34 cm.

Putting the value of x and y in the equation (2)

$z=\frac{19652}{34 \times 24}$

=17 cm.

The dimension of the cardboard is 34 cm by 34 cm by 17 cm.

4 0

3 years ago

Blank over five plus eighteen over twenty equals thirty over twenty

Elena L [17]

Answer:

3 and i use flvs too, also that pfp is fire

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A hypotanuse of a right angle triangle is 13cm if one of the side of the triangle is 1cm shorter than the hypotanuse calculate t

pentagon [3]

Answer:

(H)^2=(B)^2 +(P)^2

(13cm)^2=(12cm)^2 +(P)^2

169cm^2=144cm^2 +(P)^2

169cm^2-144cm^2 =(P)^2

25cm^2 = (P)^2

Now taking square root on both sides

5cm = P

Step-by-step explanation:

4 0

3 years ago

If d= the number of dogs ,which variable expression represents the phase below? The sun of the number of dogs and the 6 cats

Morgarella [4.7K]

Answer:

d + 6

Step-by-step explanation:

Given that :

Number of dogs = d

The sum of the number of the number of dogs and 6 cats is :

This is a straightforward question :

We take the sum of d and 6

Hence, we have

d + 6

5 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago