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vladimir1956 [14]

3 years ago

13

Suppose Ray BD bisects ZABC. If m ZABD=30 degrees what is mZDBC

1 answer:

Masteriza [31]3 years ago

7 0

Answer:

there is not enough info to determine

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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

5/12 + 7/9 can someone help me out

Montano1993 [528]

Answer:

43/36 or 1 7/36 or 1.194

4 0

3 years ago

Read 2 more answers

Which pair of numbers has the greatest common factor As that of 48 and 78 ?

Nataly_w [17]

48 and 78(try to see if its 6)

5 0

3 years ago

In ΔNOP, the measure of ∠P=90°, NO = 49 feet, and OP = 19 feet. Find the measure of ∠N to the nearest degree.

mafiozo [28]

Answer:

N=22.815≈23

Step-by-step explanation:

SIN N=19/49

N=\sin^{-1}(\frac{19}{49})

N=sin

N=22.815≈23

7 0

3 years ago

A square plate rests horizontally in a spherical container of radius 13. The plate is 1 unit above the bottom point of the conta

Ivahew [28]

Given is :

The radius of the spherical container = 13 units

A square plate rests horizontally in a spherical container and is 1 unit above the bottom point of the container. So, the distance from the center of the sphere to the center of the plate = 12 units. And the distance from the center of the sphere to the corner of the plate is 13 units.

Lets assume the side is 'x' units

Now using Pythagoras theorem, we will find the distance from the center of the plate to the corner :

$12^{2} +x^{2} =13^{2}$

$144+x^{2} =169$

$x^{2} =25$

x=5 units

So, the length of the diagonal becomes 10 units.

Now, we get an isosceles triangle with sides x,x and 10 units.

$x^{2} +x^{2} =10^{2}$

$2x^{2} =100$

$x^{2} =50$

$x=\sqrt{50}$

Hence, the length of a side of this plate is $\sqrt{50}$ units.

5 0

3 years ago