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____ [38]
2 years ago
11

Which of the following options have the same value as 30\%30%30, percent of 818181?

Mathematics
1 answer:
Nataly_w [17]2 years ago
6 0

Answer:

Option B is correct = 0.3 \times 81

Step-by-step explanation:

<u>The complete question is:</u> Which of the following options have the same value as 30% of 81?

Group of choices is:

(A) \frac{30}{100}\times 81 \times 100

(B) 0.3 \times 81

(C) 0.03 \times 81

(D) \frac{3}{10}\times 81 \times 10

(E) 30 \times 81

Now, the expression given to us is 30% of 81.

Simplifying the above expression we get;

   30% of 81  =  \frac{30}{100} \times 81

                     =  \frac{3}{10} \times 81  =  0.3 \times 81

Now, we will solve each of the given options and then see which option matches with our calculation.

Option (A) is given;

\frac{30}{100}\times 81 \times 100  =  30 \times 81

This doesn't match with our answer, so this option is not correct.

Option (B) is given;

0.3 \times 81  

<u><em>This matches with our answer, so this option is correct.</em></u>

Option (C) is given;

0.03 \times 81  

This doesn't match with our answer, so this option is not correct.

Option (D) is given;

\frac{3}{10}\times 81 \times 10  =  3 \times 81

This doesn't match with our answer, so this option is not correct.

Option (E) is given;

30 \times 81  

This doesn't match with our answer, so this option is not correct.

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artcher [175]
The fraction is 2/3
So, there should be 100 squares.
Next, the 100 squares is divided into 3 by vertical lines. The lines will not coincide with the sides of the squares. This is okay. Next, the 2 out of 3 will be shaded, this will be equivalent to 66 squares and one more square but it will not be filled up completely. In percent form, the fraction is 66.67%.
6 0
2 years ago
The volume of clay used to build a cylindrical pillar with a height of 9 centimeters is 324π cubic centimeters. The area of the
stealth61 [152]
Area=height times base (for some prisms including cylinders)
vcylinder=hpir²
h=height
therefor pir²=base area
vcylinder=height times aeraofbase

 

given
h=9
v=324pi
324pi=9(basearea)
divide both sides by 9
36pi=areabase

the area of te base is 36pi square cm (put 36 in the blank since th pi is alredy there)
3 0
2 years ago
Read 2 more answers
A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
tino4ka555 [31]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Step-by-step explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

y = 1.6 x - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y = - \frac{1.6}{2( - 0.032)} = 25

and maximum height is

y = 1.6(25) - 0.032 (25)^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

6 0
2 years ago
There are 648 cows and sheep on the farm. The ratio of the number of Cows to the number if sheep is 5:4. At an auction 2 thirds
GREYUIT [131]
Cows : sheep = 5 : 4

5x + 4x = 648

9x = 648

x = 72

5x = 5 * 72 = 360

4x = 4 * 72 = 288

There are originally 360 cows and 288 sheep.

2/3 of the sheep are sold. That means that 1/3 of the sheep are left.

1/3 * 288 = 96

96 sheep are left.

x cows were sold.

(360 - x)/96 = 3/1

360 - x = 288

-x = -72

x = 72

72 cows were sold
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2 years ago
What is the maximum number of solutions each of the following systems could have?
Marrrta [24]

Answer:

Two distinct concentric circles: 0 max solutions

Two distinct parabolas: 4 max solutions

A line and a circle: 2 max solution

A parabola and a circle: 4 max solutions

Step-by-step explanation:

<u>Two distinct concentric circles:</u>

The maximum number of solutions (intersections points) 2 distinct circles can have is 0. You can see an example of it in the first picture attached. The two circles are shown side to side for clarity, but when they will be concentric, they will have same center and they will be superimposed. So there can be ZERO max solutions for that.

<u>Two distinct parabolas:</u>

The maximum solutions (intersection points) 2 distinct parabolas can have is 4. This is shown in the second picture attached. <em>This occurs when two parabolas and in perpendicular orientation to each other. </em>

<u>A line and a circle:</u>

The maximum solutions (intersection points) a line and a circle can have is 2. See an example in the third picture attached.

<u>A parabola and a circle:</u>

The maximum solutions (intersection points) a parabola and a circle can have is 4. If the parabola is <em>compressed enough than the diameter of the circle</em>, there can be max 4 intersection points. See the fourth picture attached as an example.

4 0
2 years ago
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