All categories

3 years ago

How many factors of muliplication does 20 have

Mathematics

2 answers:

Igoryamba3 years ago

8 0

1 X 20
2 X 10
4 X 5

So actually there are 6 factors of 20

amid [387]3 years ago

4 0

Here are some factors of 20!

1x20=20

2x10=20

4x5=20

5x4=20

Therefore there are 4 factors of 20.

You might be interested in

iris charges a fee for her consulting services plus an hourly rate that is 1 1/5 times her fee on a 7 hour job and iris charged

Serggg [28]

The question is asking to calculate the fee and the hourly rate of Iris base on the data in the problem, i my own calculation and analysis, I would say that the fee of Iris is $50, and her hourly rate would be $60, I hope you are satisfied with my answer and feel free to ask for more

6 0

3 years ago

Use the quadrant system to find the area of the polygon

AfilCa [17]

Answer:

The area of the polygon is 64 units²

Step-by-step explanation:

* Lets explain how to solve the problem

- The polygon has 5 sides and 5 vertices

- To find its area by easy way split it into two trapezoid by a horizontal

line y = 1

- The vertices of trapezoid (1) are (-5 , 1) , (4 , 1) , (-5 , 5) , (0 , 5)

- The vertices of one of the parallel bases are (-5 , 1) , (4 , 1)

- The length of this base = 4 - (-5) = 4 + 5 = 9 units

- The vertices of the other parallel bases are (-5 , 5) , (0 , 5)

- The length of this base = 0 - (-5) = 0 + 5 = 5 units

- The length of its height = 5 - 1 = 4 units

- The area of the trapezoid = 1/2 (b1 + b2) × h

∴ Area of trapezoid 1 = 1/2 (9 + 5) × 4 = 28 units²

- The vertices of trapezoid (2) are (4 , 1) , (-5 , 1) , (-5 , -5) , (-2 , -5)

- The vertices of one of the parallel bases are (4 , 1) , (-5 , 1)

- The length of this base = 4 - (-5) = 4 + 5 = 9 units

- The vertices of the other parallel bases are (-5 , -5) , (-2 , -5)

- The length of this base = (-2) - (-5) = -2 + 5 = 3 units

- The length of its height = 1 - (-5) = 1 + 5 = 6 units

∴ Area of trapezoid 1 = 1/2 (9 + 3) × 6 = 36 units²

∵ Area of the polygon = Area trapezoid (1) + Area trapezoid (2)

∵ Area trapezoid (1) = 28 units²

∵ Area trapezoid (2) = 36 units²

∴ Area of polygon = 28 + 36 = 64 units²

∴ The area of the polygon is 64 units²

3 0

4 years ago

(4 - 3x) is greater than or equal to -9

dalvyx [7]

Answer:

x ≤ 13/3

Step-by-step explanation:

4 - 3x ≥ -9

Isolate the variable, x. Treat the ≥ sign as an equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 4 from both side:

4 (-4) - 3x ≥ -9 (-4)

-3x ≥ - 9 - 4

-3x ≥ -13

Next, divide -3 from both sides. Note that when you divide by a negative sign, you must flip the sign:

(-3x)/-3 ≥ (-13)/-3

x ≤ (-13)/(-3)

x ≤ 13/3

x ≤ 13/3 is your answer.

8 0

4 years ago

Which diagram is NOT a good model of 2 ÷ 18? Math item response image Math item response image Math item response image (I don't

Stells [14]

Answer:

when you wrote question there will be pin sign then you will attach any image easily

Step-by-step explanation:

8 0

3 years ago

There are three different hoses used to fill a pool: hose x, hose y, and hose z. Hose x can fill the pool in a days, hose y in b

WARRIOR [948]

Answer:

C) I and III only

Step-by-step explanation:

Let full pool is denoted by O

Days Hose x takes to fill pool O = a

Pool filled in one day x = O/a

Days Hose y takes to fill pool O = b

Pool filled in one day y = O/b

Days Hose z takes to fill pool O = c

Pool filled in one day z = O/c

It is given that

a>b>c

$a>b>c>d\\\implies x$

Days if if x+y+z fill the pool together = d

1 day if x+y+z fill the pool together $=O(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{O}{d}---(1)$

I) d < c

d are days when hose x, y, z are used together where as c are days when only z is used so number of days when three hoses are used together must be less than c when only z hose is used. So d < c

III) $\frac{c}{3}$

Using (1)

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad(a>b>c)\\(ab+bc+ca)\frac{abc}{3ab}\\\\d>\frac{c}{3}$

Similarly

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad a>b>c\\(ab+bc+ca)>3bc\\\\d=\frac{abc}{ab+bc+ca}$

So,

$\frac{c}{3}$

3 0

3 years ago