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Yuri [45]
3 years ago
8

I don't know how to answer number 9 and 10. Can someone please help me?

Mathematics
1 answer:
kogti [31]3 years ago
5 0

AB = AC = 32

SR = TR = 7

Solution:

Question 9:

<u>Tangent theorem of a circle:</u>

Lengths of tangents drawn from an same external point to a circle are equal.

In figure AB and AC are tangents drawn from same an external point A.

By tangent theorem of a circle,

⇒ AB = AC

\Rightarrow 2x^2=8x

Divide both side of the equation by x.

$\Rightarrow \frac{2x^2}{x} =\frac{8x}{x}

$\Rightarrow 2x =8

Divide by 2 on both side of the equation.

$\Rightarrow \frac{2x}{2} =\frac{8}{2}

$\Rightarrow x =4

Substitute x = 4 in AB and AC.

AB=2(4)^2=32

AC=8(4)=32

Question 10:

By tangent theorem of a circle,

⇒ SR = TR

$\Rightarrow y=\frac{y^2}{7}

Multiply by 7 on both side of the equation, we get

$\Rightarrow 7y =y^2

Divide by y on both side of the equation, we get

$\Rightarrow 7 =y

Substitute y = 7 in SR and TR

SR=y=7

$TR=\frac{7^2}{7} =7

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You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3
Charra [1.4K]

Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

h=\frac{504}{lw}

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides C_{s}= 3[2(l + w)h] = 6(l + w)h

C_{s}= 3[2(l + w)h]

C_{s}=6(l+w)(\frac{504}{lw} )

Cost of the top and the bottom C_{(t,p)}= 4(2lw) = 8lw

Total cost of the box C = 3024\frac{(l+w)}{lw} + 8lw

                                      = 3024[\frac{1}{l}+\frac{1}{w}] + 8lw

To minimize the cost of the sides

\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0

\frac{3024}{l^{2}}=8w

\frac{378}{l^{2}}=w ---------(1)

\frac{dC}{dw}=3024(-w^{-2})+8l=0

\frac{3024}{w^{2}}=8l

\frac{378}{w^{2}}=l

w^{2}=\frac{378}{l}-------(2)

Now place the value of w from equation (1) to equation (2)

(\frac{378}{l^{2}})^{2}=\frac{378}{l}

\frac{(378)^{2} }{l^{4}}=\frac{378}{l}

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

w^{2}=\frac{378}{7.23}

w^{2}=52.28

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

h=\frac{504}{(7.23)^{2}}

h = 9.64 cm                        

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