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3 years ago

7

Help - i'll give brainliest

2 answers:

Serggg [28]3 years ago

8 0

Answer:

W=91.2

Step-by-step explanation:

kherson [118]3 years ago

4 0

7.6=1212⋅7.6=12⋅12

=91.2

have a good day

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Can someone help me with this ?

vesna_86 [32]

24+56= 80
24/80= .30

30%

4 0

3 years ago

You have to use the numbers 1,2,3,4 and 5. Add those numbers and the sum should be equal to 50. You can combine the numbers to f

victus00 [196]

I think we must use two numbers which have two digits and one of them must be one digited number.If we try it we can say our numbers are xy,ab and c.

xy + ab + c = 50

10x+y+10a+b+c = 50

10(x+a) + y+b+c = 50

Think about y+b+c. Maximum it can be 5+4+3 = 12 but it can't because there is a number which is multiple of 10.So it can be 10 maximum and x+a must be 4.

We can only use 1+3. 40 + y+b+c = 50 , y+b+c = 10. And other numbers are 2+4+5 = 11. 10 doesn't equal to 11 and it says there isn't any solution for this problem :)

4 0

3 years ago

Read 2 more answers

A man who was outside in the rain without an umbrella or hat didn’t get a single hair on his head wet. Why?

aksik [14]

Answer:

perhaps he was bald

Step-by-step explanation:

if hes bald, no hair can get wet

8 0

3 years ago

Read 2 more answers

<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

Please help its due tomorrow

vekshin1

A) - 4/6
B)- 1/6
D 1 1/3

3 0

3 years ago