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Illusion [34]
3 years ago
12

How do you tell whether the percent of a number will be greater than, less than, or equal too the number

Mathematics
1 answer:
Ulleksa [173]3 years ago
8 0
If the tens are greater for one number then that number will be greater. Parent Note: ... The tens are equal so we need to compare the ones.
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Is 1.572735 rational?
Gekata [30.6K]

Answer:

yes

Step-by-step explanation:

it is rational

6 0
3 years ago
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A shirt that normally costs $35.00 is on sale for 28.00. What is the percent discount?
Olin [163]

Answer:

$25.20

Step-by-step explanation:

$35 x 0.28

= $9.80

35-9.80

= $25.20

3 0
3 years ago
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Please help im offering the highest amount of points I can give.
STatiana [176]

Answer:

96 cm^3

Step-by-step explanation:

I hope this helped!

5 0
3 years ago
Solve each system using substitution or additions<br> (elimination).<br><br> -x+2y=-1<br> 5x-10y=6
NNADVOKAT [17]

Answer:

Step-by-step explanation:

L1 -x + 2y = -1

L2 5x - 10y = 6

-x + 2y = - 1 multiply by -1

x - 2y = 1 multiply by 5

5x - 10y = 5 is similar that 5x - 10y = 6

The system has no solution

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
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