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3 years ago

Please Please help me with this answer

Mathematics

2 answers:

tatyana61 [14]3 years ago

4 0

3 bowls of chicken

7 bowls of tomato.

3+7=10

7-3=4

bagirrra123 [75]3 years ago

3 0

Let's give our numbers names: we'll call the number of bowls of tomato soup t and the number of bowls of chicken soup c.

We know from the question that c is 4 less than t, so we can write that fact as the equation

c = t - 4

We also know that, when we add both of the amounts together, we get 10, so that

c + t = 10

We can then use the first equation to replace c with t - 4:

(t - 4) + t = 10

or, rearranging:

2t - 4 = 10

This means that if we double the number of tomato bowls and subtract 4, we arrive at 10. To find what exactly t is, then, we can go the other way around. If we add 4, we see that 2t = 14, and if we halve 2t, we find that t = 7, so the cafeteria sold 7 bowls of tomato soup.

Checking our work, we see that c = 7 - 4 = 3, and 7 + 3 = 10, so this answer is consistent with what we've been given.

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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

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If three angles of a hexagon are each x° and each of the other is 1/2 times x° find the angles

Elena-2011 [213]

Answer:

Step-by-step explanation:

The given hexagon has interior angles

x,x,x, x/2, x/2, x/2 for a total of S1=4.5x.

We also know that the sum of interior angles of any polygon of n sides is

S2=180(n-2)

For a hexagon, S2 = 180(6-2) = 720°.

Equating S1 and S2

4.5x = 720

x = 720/4.5 = 160.

Therefore substitute x in the given angles to get

{160°,160°.160°,80°,80°,80°}

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Step-by-step explanation:

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