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wariber [46]
3 years ago
10

How do I solve (2-3*2-3)-4

Mathematics
2 answers:
Anestetic [448]3 years ago
8 0
Do 2-3 times 2-3 then -4
kondor19780726 [428]3 years ago
3 0
According to PEMDAS, parenthesis goes first. Before addition, multiplication is represented in the M of PEMDAS. So you multiply 3 and 2 inside of the parenthesis. That equals to 6. (2-6-3)-4 is what is left. Then, we still have subtraction to be done in the parenthesis. 2 minus 6 is -4, minus 3 is -7. You’re left with -7-4. This equals to -11. Your answer is -11.
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Answer:

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Step-by-step explanation:

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What must be the sides of an equilateral triangle so that it's area should be equal to the area of an isosceles triangle with ba
d1i1m1o1n [39]

Answer:

The sides of the equilateral triangle are 11.7 m.

Step-by-step explanation:

Let's find the area of the isosceles triangle:

A_{i} = \frac{bh}{2}

Where:

b: is the base = 12 m

h: is the height

We can find the height by using Pitagoras:

x^{2} = \frac{b^{2}}{2} + h^{2}    

Where:

x is the hypotenuse = side of the triangle = 10 m

h = \sqrt{x^{2} - \frac{b^{2}}{2}} = \sqrt{(10)^{2} - 6^{2}} = 8 m

Then, the area is:

A_{i} = \frac{12*8}{2} = 48 m^{2}

Now, since the area of the isosceles triangle is equal to the area of the equilateral triangle:

A_{e} = A_{i} = 48 m^{2}

A_{e} = \frac{bh}{2}

The height of the equilateral triangle is given by:

b^{2} = \frac{b^{2}}{2} + h^{2}    

h = \sqrt{b^{2} - \frac{b^{2}}{2}} = \frac{b}{\sqrt{2}}

Hence, the sides are:

A_{e} = \frac{1}{2}b\frac{b}{\sqrt{2}} = \frac{b^{2}}{2\sqrt{2}}            

b = \sqrt{A*2\sqrt{2}} = \sqrt{48*2\sqrt{2}} = 11.7 m      

Therefore, the sides of the equilateral triangle are 11.7 m.

I hope it helps you!      

7 0
3 years ago
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Coleen, had a choise between 3/10 of a bag of candy or 3/6 of a bag of candy.if she want to get as much candy as possible, which
Zina [86]

Answer:

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Step-by-step explanation:

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So, 6x3=18 and 10x3=30

and 3/6 has a higher value, so its correct

3 0
4 years ago
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