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DENIUS [597]
2 years ago
9

Can someone please give ,e a good explanation of associative property and examples please​

Mathematics
1 answer:
Bess [88]2 years ago
6 0

Answer:

See below

Step-by-step explanation:

Basically, the associative property states that when an expression has three terms, they can be grouped in any way to solve that expression.

For example, (8*4)*5=8*(4*5) if you're doing multiplication, or (8+4)+5=8+(4+5), if you're doing addition.

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3x +2y=6 solve for Y
ololo11 [35]

Answer:

y = \frac{6-3x}{2}

Step-by-step explanation:

Given

3x + 2y = 6 ( subtract 3x from both sides )

2y = 6 - 3x ( divide both sides by 2 )

y = \frac{6-3x}{2}

8 0
3 years ago
What is the area of the figure below? PLEASE HELP ​
viktelen [127]

Answer:

Step-by-step explanation:

This can be broken down into a square and a rectangle.

A=5(5)+(12)(8-5)

A=25+12(3)

A=25+36

A=61 u^2

6 0
3 years ago
Read 2 more answers
I need help with this question I would be grateful really
andreyandreev [35.5K]

Answer:

B

Step-by-step explanation:

B

the vertex is (-5,-2) which eliminates A and C. D says that the x intercepts are both positive which is obviously incorrect making B the correct answer

6 0
3 years ago
If 16 oz equal 1 pound how much ounces are equal to 3 pounds
Solnce55 [7]

Answer:

it would be 48oz

Step-by-step explanation:

16 times 3 = 48oz -sorry if this is not right

have a wonderful day :)

3 0
2 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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