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kiruha [24]
4 years ago
10

Is 5622 divisible by 4

Mathematics
2 answers:
Pani-rosa [81]4 years ago
7 0
Well it's 1,405.5 so no
frez [133]4 years ago
3 0

Actually no because when we divide 5622/4 = 2811/2. So it is not divisible.
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olga_2 [115]

Answer:

Domain → (-∞, ∞)

Range → (-2, ∞)

Step-by-step explanation:

1). Domain of a function is defined by the input values (set of x-values) and Range of the function is defined by the output values (set of y-values).

From the graph attached,

Function is defined for all real values of x.

Therefore, domain of the function will be (-∞, ∞).

On y-axis values of the function vary from -2 (excluding 2) to positive infinity.

Therefore, range of the function will be (-2, ∞).

2). Average rate of change of a function in the interval (a, b) is defined by,

Average rate of change = \frac{f(b)-f(a)}{b-a}

By using this expression we can find the average rate of change in the given interval.

Please give the correct interval for which the average rate of change is to be calculated.

8 0
3 years ago
Please help I forgot how to do this
iren2701 [21]

Answer:

56.

Step-by-step explanation:

Formula for solving a triangle is 1/2 * base * height.

3 0
3 years ago
If a cube of metal measures 3 cm on each side and has a mass of 216 g what is the density of the metal
ohaa [14]
The formula of density is mass / volume.
So, the volume is equal to the area of the cube by the total height, so the area of the cube is 3x3= 9, and this result multiplied by the height, 9x3 = 27.
Following the formula of density, we have 216 g / 27 cm^3, the result is 8 g/cm^3.
Hope you undestand my procedure XD
4 0
4 years ago
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What is -3x^2-12x-10 in vertex form??
Alla [95]
If you type the equation on math way it will show you a graph for your answer!
5 0
3 years ago
The members of a high school band asked a number of students whether they would like blue, gold, or both for the uniforms for th
MaRussiya [10]

The values of a and b are a = 33% and b = 43% ⇒ last answer

Step-by-step explanation:

The venn-diagram contains:

  • A circle labeled blue 32
  • A circle labeled gold 25
  • The two circles are overlap with label 12 on the overlap

The table:

→ Band  :  blue  :  N.b  : total

→ gold   :  16%   :  a      :  49%

→ N.g     :  b       :  8%   :  51%

→ Total  :  59%  :  41%  :  100%

We need to find the values of a and b

From the table

16% represents the percentage of blue and gold uniforms (2nd row with 2nd column)

From venn-diagram

The common part of blue and gold label with 12 (overlap of two circles)

∴ 16% of the total number of students = 12

∵ 16% × x = 12

∴ \frac{16}{100} × x = 12

∴ 0.16 x = 12

- Divide both sides by 0.16

∴ x = 75

∵ x represents the total number of students

∴ The total number of students were asked is 75

From the table

b represents the percentage of blue but not gold uniforms (3rd row and 2nd column)

From venn-diagram

The part that has blue and not gold labeled with 32 (the part of the blue circle not in the gold circle)

∴ 32 of the total number of students like the blue but not gold

- Divide 32 by the total 75 , then multiply the quotient by 100%

   to find the percentage of blue but not gold

∴ b = \frac{32}{75} × 100% = 42.6666%

∴ b ≅ 43%

From the table

a represents the percentage of gold nut not blue uniforms (2nd row and 3rd column)

From venn-diagram

The part that has gold and not blue labeled with 25 (the part of the gold circle not in the blue circle)

∴ 25 of the total number of students like the gold but not blue

- Divide 25 by the total 75 , then multiply the quotient by 100%

   to find the percentage of gold but not blue

∴ a = \frac{25}{75} × 100% = 33.3333%

∴ a ≅ 33%

The values of a and b are a = 33% and b = 43%

To check your answer complete the table you will find the total of it is 100% ( for the columns 16% + 43% = 59% , 33% + 8% = 41% , then 59% + 41% = 100%, for the rows 16% + 33% = 49% , 43% + 8% = 51% , then 49% + 51% = 100%)

Learn more:

You can learn more about the probability in brainly.com/question/3756853

#LearnwithBrainly

4 0
3 years ago
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