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marysya [2.9K]
3 years ago
9

Which equation demonstrates the multiplicative identity property?

Mathematics
1 answer:
12345 [234]3 years ago
4 0

Answer:

(-3+5)(1)=-3+5

Step-by-step explanation:

The multiplicative identity is 1.

The multiplicative identity property days that, if you multiply a real number by 1, that result is the same real number.

This is demonstrated in option B of the given alternatives, where we have:

( - 3 + 5)(1) =  - 3 + 5

The multiplication by did not change b the value.

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Combine like terms to create an equivalent expression.
Brut [27]

Question..

Combine like terms to create an equivalent expression.

½ −⅙q +⅚q - ⅓

Answer:

½ −⅙q +⅚q - ⅓ is equivalent to ⅔q + ⅙

Step-by-step explanation:

Given

½ −⅙q +⅚q - ⅓

Required

Equivalence

½ −⅙q +⅚q - ⅓

We start by collecting like terms.

⅚q - ⅙q + ½ - ⅓

Factorize

(⅚ - ⅙)q + ½ - ⅓

((5 - 1)/6)q + ½ - ⅓

(4/6)q + ½ - ⅓

Reduce 4/6 to lowest term

⅔q + ½ - ⅓

Evaluate fraction

⅔q + (3 - 2)/6

⅔q + ⅙

Hence, ½ −⅙q +⅚q - ⅓ is equivalent to ⅔q + ⅙

7 0
3 years ago
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This Line Chart shows the number of soft drink bottles a vendor sold during each month of baseball season.
navik [9.2K]
The answer is A: 50



400 - 350 = 50
4 0
2 years ago
What is 6.84 in word form
alexandr402 [8]
Six point eight four.

(Note, you don't say, "Six point eighty-four" As it doesn't sound correct.

Cheers.
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3 years ago
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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 }) }

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3 years ago
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Please help me<br><br><br> Simplify.<br><br> 3(x + 4)
valentinak56 [21]
Distribute the 3.

3*x + 3*4
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Hope this helps :)
8 0
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