Pouch-and-coin situations

Mathematics

1 answer:

ser-zykov [4K]3 years ago

4 0

What are you talking about if you have a question ask me

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Can u help me on this?

a_sh-v [17]

Answer:

Step-by-step explanation:

70-20=50

i don't sure correct or not

7 0

3 years ago

??????????????????????????

Umnica [9.8K]

Solution:

Simple Interest = Principal Amount × Rate of Interest/100 × Time

Here, Principal Amount = $6000

Rate of Interest = 6%

Time = 4 years

Simple Interest = 6000 × 6/100 × 4 = $1440

3 0

3 years ago

Which of the following fractions is smaller than 1/2

Harman [31]

the answer is C: 3/7

5 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

7 0

3 years ago

Answer?I need it pleaseeee

Elan Coil [88]

Answer: 20 dallors

Step-by-step explanation:

8 0

2 years ago