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RoseWind [281]
3 years ago
5

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:

B

Step-by-step explanation:

70-20=50

i don't sure correct or not

7 0
3 years ago
??????????????????????????
Umnica [9.8K]

Solution:

<em>Simple Interest = Principal Amount × Rate of Interest/100 × Time</em>

Here, Principal Amount = $6000

Rate of Interest =  6%

Time = 4 years

Simple Interest = 6000 × 6/100 × 4 = <em>$1440</em>

3 0
3 years ago
Which of the following fractions is smaller than<br> 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 <span>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.
</span>
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
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