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Anastasy [175]
3 years ago
13

How do you answer this ??

Mathematics
1 answer:
mel-nik [20]3 years ago
5 0
This is how you do it. Set two equations equal to each other then solve. Plug in x for each equation, find the answers. AC is the base because it is not congruent to any other side.

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Should montresor be tried for murder?why, or why not?Does he have a case for an insanity plea?explain.​
vaieri [72.5K]

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Montresor deserves to be penalised in court with the sentencing of 1st degree murder. Montresor was not insane, he was a pervasive murderer because of his motive.

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--Applepi101

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3 years ago
Which area model shows the product of
adell [148]
A it is right i know it is becous i did this yester day
4 0
3 years ago
Read 2 more answers
O 91.7 miles<br> O 55.5 miles<br> O 41.7 miles<br> O 63.8 miles
Art [367]

Answer:

* 55.5 miles

Step-by-step explanation:

17² + b² = 23²

289 + b² = 529

b² = 240

b = 15.491 ≈ 15.5

17 + 23 + 15.5 = 55.5

You can also simply notice B is the only answer with a .5 decimal and thus don't have to add.

4 0
3 years ago
WILL GIVE BRAINLIEST PLZ HURRY
Gala2k [10]

Answer:

Joey has 3 pets.

Step-by-step explanation:

f. 3x = 9

9 divided by 3 is 3.

x = 3

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4 0
3 years ago
A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Mater
frutty [35]

Answer:

Cost of material is $245.31

Step-by-step explanation:

Let dimension of box,

  • Length, l = x m
  • Width, w = y m
  • Height, h = z m

<em>The length of this base is twice the width.</em>

<em>∴ </em>x = 2y

Volume of box = 10 m³

∴ xyz = 10

⇒ 2y²z = 10

⇒ y²z = 5

\Rightarrow z=\dfrac{5}{y^2}\ \ \ ...(i)

<em>Material for the base costs $15 per square meter.</em>

Total cost of base = 15xy

Total cost of base = 30y²         [∵ x = 2y ]

<em>Material for the sides costs $9 per square meter.</em>

Total cost of side = 9(2xz+2zy)

Total cost of side = 18(xz+yz)

Total cost of material for container = 30y² + 18(xz+yz)

C(y)=30y^2+18(2y\cdot \dfrac{5}{y^2}+y \dfrac{5}{y^2})     [From (i)]

C(y)=30y^2+\dfrac{270}{y}

Differentiate w.r.t y

C'(y)=60y-\dfrac{270}{y^2}

For critical point , C'(y)=0

60y-\dfrac{270}{y^2}=

y=\sqrt[3]{\dfrac{9}{2}}

x=2y=2\sqrt[3]{\dfrac{9}{2}}

z=\dfrac{5}{y^2}=5\sqrt[3]{\dfrac{4}{81}}

The minimum cost of container material  at y=\sqrt[3]{\dfrac{9}{2}}

C_{min}=30\sqrt[3]{\dfrac{81}{4}}+270\sqrt[3]{\dfrac{2}{9}}

C_{min}=245.31

Hence, the cheapest cost of material for container is $245.31

6 0
3 years ago
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