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Kitty [74]
3 years ago
5

Which of the following shows the correct process for finding the area of the shape shown below?​

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
7 0

Answer:

A = bh = A = 9.5*14

Step-by-step explanation:

Tems11 [23]3 years ago
6 0
What they said ⟟ think
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Sandra has ribbons that are 2/6 meter 3/4meter and 4/9 meter long .She needs to use the ribbon longer than 2/3meter to make a bo
timurjin [86]

Answer:

3/4 meter

Step-by-step explanation:

Need Common denominators to compare size

2/6  3/4  4/9     which is longer than 2/3?

2/6 compared to 2/3.  2/6 = 2/6 and 2/3 = 4/6.  

2/6 < 2/3 so 2/6 is too short

3/4 compared to 2/3   3/4 = 9/12  and 2/3 =  8/12

3/4 > 2/3 so 3/4 is longer than 2/3

4/9 compared to 2/3  4/9 = 4/9 and 2/3 = 6/9

4/9 < 2/3 so 4/9 is too short

6 0
3 years ago
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4 years ago
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What is this equal<br> how can I solve similar trigonometric integrals like this one
Angelina_Jolie [31]

Answer:

ln|sec θ + tan θ| + C

Step-by-step explanation:

The integrals of basic trig functions are:

∫ sin θ dθ = -cos θ + C

∫ cos θ dθ = sin θ + C

∫ csc θ dθ = -ln|csc θ + cot θ| + C

∫ sec θ dθ = ln|sec θ + tan θ| + C

∫ tan θ dθ = -ln|cos θ| + C

∫ cot θ dθ = ln|sin θ| + C

The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.

∫ sec θ dθ

∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ

∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ

ln|sec θ + tan θ| + C

3 0
3 years ago
The base of an aquarium with given volume is made of slate and the sides are made of glass. If slate costs five times as much (p
crimeas [40]

Answer:

Base = Length = (2V/5)^⅓

Height = V^⅓/(2/5)^⅔

Step-by-step explanation:

Let the sides of the aquarium be l,b and h

Where l = length

b = base

h = height

Volume = lbh

V = lbh

The surface area of the box is

A = 2bh + 2lh + 2lb

We'll replace the 2lb with 5lb to get the cost of area because it's given the question that the slate (base) of the box cost 5 times as much per unit area as glass

C(l,b,h) = 2bh + 2lh + 5lb

Make h the subject of formula in (V = lbh)

h = V/lb

This will enable us to account for the lowest cost of materials by taking derivatives of the cost equation, and we need to solve for a l to put into the solution.

Substitute V/lb for h in the cost equation

C(l,b,V/lb) = 2b(V/lb) + 2l(V/lb) + 5lb ------ Simplify

C(l,b,V/lb) = 2V/l + 2V/b + 5lb

Take the derivatives of the above with respect to l and b

Cl = -2V/l² + 5b

Cb = -2V/b² + 5l

Equate Cb to Cl (this implies that b = l)

Cb = Cl =>

-2V/l² + 5b = -2V/b² + 5l

So, we have

C(b,b) = -2V/b² + 5b = 0

-2V/b² + 5b = 0 ---- Solve for b

-2V/b² = -5b ---- Multiply through by b²

-2V = -5b³ ---- Divide through by -5

2V/5 = b³ ---; Rearrange

b³ = 2V/5

b = (2V/5)^⅓

b = l

So, l = (2V/5)^⅓

h = V/lb ---- (b = l)

h = V/l² ---- Substitute (2V/5)^⅓ for l

h = V/((2V/5)^⅓)²

h = V/(2V/5)^⅔

h = V/((V^⅔)(2/5)^⅔)

h = V^⅓/(2/5)^⅔

5 0
4 years ago
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