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lozanna [386]
3 years ago
15

MATH QUESTION : As punishment for bad behavior during recess, Mrs. Busywork asked her class to multiply 10 by 1/3 five times. Jo

hn, however, notices that it is possible to multiply 10 by a single fraction and still get the same answer as the other students. What is this single fraction?
Mathematics
1 answer:
Rom4ik [11]3 years ago
6 0

Answer:

5/3

Step-by-step explanation:

5×(10×1/3)

=10×5/3

=since 5×(10×1/3) gives 50/3 so thus 10×5/3.

The fraction is therefore 5/3

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Vadim26 [7]

Answer:     5 pls44444444 do you want my ps5

Step-by-step explanation:

8 0
3 years ago
14.
Charra [1.4K]

a. The number of square cm of paper that would be used to make 4 caps is 3154.56 cm².

b. For the cake to have the required dimensions, they should order 1 kg of cake.

a.

The number of square cm of paper that would be used to make 4 caps is 3154.56 cm².

To find the number of square cm paper would be used to make 4 caps, we need to find the area of a cone.

<h3>Area of a cone</h3>

The area of a cone, A = πr² + πr[√(h² + r²)] where

  • r = radius of cone and
  • h = height of cone

Now given that the base circumference of the cone is C = 44 cm. So,

C = 2πr where r = radius of base

So, r = C/2π = 44 cm/2π = 22 cm/3.142 = 7 cm

The height of the cone, h = 24 cm

So, A = πr² + πr[√(h² + r²)]

A = π(7 cm)² + π(7 cm)[√((24 cm)² + (7 cm)²)]

A = π(7 cm)² + π(7 cm)(7)[√((4 cm)² + (1 cm)²)]

A = 49π cm² + 49π cm[√(16 cm² + 1 cm²)]

A = 49π cm² + 49π cm[√(17 cm²)]

A = 49π [1 cm² + 4.123 cm²)]

A = 49π [5.123 cm²)]

A = 251.032π

A = 788.64 cm²

Since the area of one cap is A = 788.64 cm²,

the area of 4 caps is A' = 4A

= 4 × 788.64 cm²

= 3154.56 cm²

So, the number of square cm of paper that would be used to make 4 caps is 3154.56 cm².

b.

For the cake to have the required dimensions, they should order 1 kg of cake.

To find the amount of cake to be ordered, we need to find the volume of a cake since it is the volume of a cylinder.

<h3>Volume of a cylinder</h3>

The volume of a cylinder is V = πd²h/4 where

  • d = diameter of cake = 24 cm and
  • h = height of cake = 14 cm

Substituting the values of the variables into the equation, we have

V = πd²h/4

V = π(24 cm)²(14 cm)/4

V = π(576 cm)²(14 cm)/4

V = 8064π cm³/4

V = 2016π cm³

V = 6333.45 cm³

So the volume of the cake is 6333.45 cm³

Since 650 cm³ equals 100 g, 6333.45 cm³ = 6333.45 cm³ × 100 g/650 cm =  974.4 g = 0.974 kg ≅ 1 kg.

So, for the cake to have the required dimensions, they should order 1 kg of cake.

Learn more about volume of a cake here:

brainly.com/question/27570775

#SPJ1

5 0
2 years ago
I need to know how to answer this question
kvv77 [185]
For the first one the answer is 17
34÷2=17
4 0
4 years ago
A fire station is to be located along a road of length A, A &lt; q. If fires occur at points uniformly chosen on (0, A), where s
Shtirlitz [24]

Answer:

Step-by-step explanation:

Given that X is uniform in the interval (0,A)

X is continuous since it represents the distance

A fire station is to be located along a road of length A, A < q. If fires occur at points uniformly chosen on (0, A), we  should find where the station be located so as to minimize the expected distance from the fire

Distance can be taken as absolute values here as either side is the same.

E(|x-a|] is to be minimum

E(|x-a|)=E(x-a) for 0<x<a

         =E(a-x), for a<x<A

Using integral we find this value

E(|x-a|)=\int\limits^a_0 {x-a} \, dx +=\int\limits^A_a -{x-a} \, dx\\f(a)=\frac{a^2}{2} -\frac{(A-a)^2}{2}

f(a)=\frac{a^2}{2} +\frac{(A-a)^2}{2} \\f'(a) = a-(A-a)\\f"(a) =2

Using calculus we find that f" is positive so when f' =0 we get solution

f'(a) =0 gives

a=\frac{A}{2}

Hence fire station to be located at the mid point of 0 and A

8 0
4 years ago
Can you solely rely on induction to prove that your conclusion is correct
ch4aika [34]
You can but it isnt really a good idea to do that if you want full credit
6 0
3 years ago
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