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mestny [16]
4 years ago
13

Simplify the expression and show ALL your steps. 3y^-2*4y^5

Mathematics
1 answer:
Anit [1.1K]4 years ago
8 0
<span>Step  1 </span><span>  :</span><span> (3 • (y2)) • 22y5 </span><span>
step  2  :</span><span> 3y2 • 22y5
</span><span>Step  3  :</span>Multiplying exponential expressions
:3.1    y2 multiplied by y5 = y(2 + 5) = y7
Final result : (3•22y7<span>)
</span>hope it helps
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Greeley [361]
A) 9.06 is the square root of 82
b) a. he evaluated 82/2 instead of evaluating the square root of 82
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4 years ago
Please answer as soon as possible!!! Thank you
True [87]

Answer:

5,0 is the answer of ur question

6 0
3 years ago
4p - 5 = 19 i need the steps to solve it
stiks02 [169]

Answer:

Hi, there your answer will be p=6

Step-by-step explanation:

Step 1: Add 5 to both sides.

4p−5+5=19+5

4p=24

Step 2: Divide both sides by 4.

\frac{4p}{4} =\frac{24}{4}

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7 0
3 years ago
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A study has shown that the daily amount of milk produced by a dairy cow is
galina1969 [7]

Answer: 4.8 gallons to 7.6 gallons (appex)

Step-by-step explanation:

8 0
3 years ago
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<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
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