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3 years ago

Is a + b = b + a a commutative or associative property of addition

Mathematics

1 answer:

myrzilka [38]3 years ago

5 0

Your answer is above..........................

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The expression sin 80 cos 70 + cos 80 sin 70 is equivalent to

vekshin1

Sin(70+80) = sin 150.......................

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2 years ago

540 is 10 Times as much as 5,400

Sergio039 [100]

Yes, that is true.....

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3 years ago

Read 2 more answers

ILL MARK AS BRAINLIEST!! AND GIVE 10 POINTS!!! Are the following lines parallel, perpendicular, or neither?

Licemer1 [7]

Answer:

b perpendicular

Step-by-step explanation:

all that is needed to do is the Y2 - y1 / X2 - X1 formula and then compare the two slopes making sure they are inverse opposites

Line 1) -7 -8 / -3 -2 X= 3X

Line 2) -3 - 0 / 12 - 3 X = -1 / 3X

Because they are inverse opposites of each other they are perpendicular had they had the exact same slopes they would have been parallels.

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

Solve this equation with variables on both sides- 8x-19=-3x-41 Show work

Nastasia [14]

8x-19=-3x-41
+3x +3x
11x-19=-41
+19 +19
11x=-22
÷11 ÷11
x=-2

8 0

3 years ago