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allsm [11]
3 years ago
15

What's a 100-38+9381-921+9100-38+9381-921+9201+9381:2​

Mathematics
2 answers:
Soloha48 [4]3 years ago
8 0
<h2>Answer:</h2><h2>44626 : 2</h2><h2></h2><h2>Hope this helps!!</h2>

Sholpan [36]3 years ago
3 0

Answer:n

Step-by-step explanation:

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What is the surface area of this rectangular prism?
Art [367]
The correct answer is C.88 ft ^2
8 0
3 years ago
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Find f(–2) for the function f(x) = 3x2 – 2x 7. −13 −1 1 23
pickupchik [31]

Answer:

16

Step-by-step explanation:

f ( x ) = 3x² - 2x

To find : f ( - 2 )

f ( - 2 ) = f ( x )

Here,

x = - 2

f ( - 2 ) = 3 ( - 2 )² - 2 ( - 2 )

           = 3 ( 4 ) - 2*( - 2 )

           = 3*4 + 4

           = 12 + 4

f ( - 2 ) = 16

8 0
2 years ago
Simplify the polynomial by combining like terms
Veseljchak [2.6K]

Answer:

22 x^2

Step-by-step explanation:

Simplify the following:

11 x^2 + 11 x^2

Hint: | Add like terms in 11 x^2 + 11 x^2.

11 x^2 + 11 x^2 = 22 x^2:

Answer:  22 x^2

none of your answers are correct.

8 0
3 years ago
If angle 1 = angle 2 and angle 2 = angle 4, what property allows you to state that angle 1 = angle 4
Gnesinka [82]

Answer:

Transitive Property

Step-by-step explanation:

You can use it for example: a = b. b = c. So a = c (Transitive Property).

6 0
4 years ago
A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.
mezya [45]
1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
A=\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2}  )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 2 \pi r=x

so r= \frac{x}{2 \pi }

Area of circle = \pi  r^{2}= \pi  ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2}  }* x^{2} = \frac{1}{4 \pi } x^{2}

4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

4 0
4 years ago
Read 2 more answers
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