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3 years ago

13

Miguel spent $6 on Saturday morning and another $9 on Saturday afternoon. How much less money did he have at the end of the day

than at the beginning? Use integer addition to determine your answer .

1 answer:

stiv31 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

0=

-6

-9=

-6+-9=

-15

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A) If the 5th term of an A.P. is double the 7th term, show that the sum of the 17 terms zero.

gayaneshka [121]

Answer:

see explanation

Step-by-step explanation:

The nth term of an AP is

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where a₁ is the first term and d the common difference

Given a₅ is double a₇ , then

a₁ + 4d = 2(a₁ + 6d) , that is

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4d = a₁ + 12d ( subtract 12d from both sides )

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The sum of n terms of an AP is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ] , substitute values

$S_{17}$ = $\frac{17}{2}$ ( 2(- 8d) + 16d)

= 8.5(- 16d + 16d)

= 8.5 × 0

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2 years ago

How are polynomials like and unlike monomials? Provide an example of a monomial and an example of a polynomial. How does the fol

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3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

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3 years ago

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