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3 years ago

14

Convert into standard notation

2 answers:

kobusy [5.1K]3 years ago

5 0

Answer:

9) 3924000000000000

10) 92000000000000000

11) 0.004391

12) 0.0006825

Misha Larkins [42]3 years ago

4 0

Answer:

I have no idea waht your talking about.

Please explain.

Step-by-step explanation:

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Solve 2^3x=172 Use logarithmic please

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4 years ago

Sketch a graph of x + y= -5

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Answer:

Slope:-1

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3 years ago

Point F is the midpoint of segment DE. DF = 2x+6 and DE = 2x+36. Find DE.

tensa zangetsu [6.8K]

Answer: It already tells you what DE= 60

Step-by-step explanation:

Since F is the midpoint of DE that means DF=FE

So create equation $2x+6+2x+6=4x+12$

When DF and FE is added your answer should equal DE

So you write the equation $4x+12=2x+36$ to find x and once x is found substitute your answer for x

$4x+12(-12)=2x+36(-12)$ subtract 12 on both sides 36-12= 24

$4x=2x+24$ subtract 2x on both sides 4x-2x=2x

$2x=24$ divide 2 into both sides 24/2=12 x=12

$4(12)+12=2(12)+36$

4x12=48+12=60 2x12=24+36=60

5 0

3 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