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gladu [14]
3 years ago
7

Can someone please help

Mathematics
2 answers:
Dmitrij [34]3 years ago
8 0

Answer:

1) a. 15ft

b. 6ft

c. 30ft

2) a. 16m

b. 12m

c. 192m

Solnce55 [7]3 years ago
7 0

answer for the 1st question:

1) length is 15 ft

2) width is 6ft

3) the area is 90 ft²

step-by-step explanation:

ok so we know that the plan is 2 in/6 ft

step 1-

we need to find the length of the terrace, and we know that the length of the terrace in the plan is 5 in

so by proportion- we'll need to find the length of the 'actual' terrace

2 in/ 6 ft= 5 in/ x ft

x=6*5/2

x=15 ft (we now have the length)

step 2-

now that we know the length, we need to find the width

the length of the terrace in the plan is 2 in

and by proportion- we'll find the 'actual' width of the terrace

2 in/6 ft=2 in/ x ft

x=6*2/2

x=6 ft

and now we'll find the area of the terrace (step 3)-

the formula for the area of a rectangle is a=lw or area=length*width

now we just fill the variables in with what we know-

so we know that 15 ft is the length and that 6 ft is the width-

15*6=90 ft²

**sorry about my original answer i didn't pay close enough attention**

good luck :)

i hope this helps

have a nice day !!

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OLga [1]

Answer: 16

Step-by-step explanation: we will do PEMDAS for this. first 6-3= 3    this makes the equation 4+8/2*3    

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3 years ago
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Ah! Okay, need help solving 14, and just checking for 16 and 4.
Allisa [31]
4. Correct. You also could have used the limit test for divergence for the same conclusion (the summand approaches infinity).

- - -

14. I'm guessing the instructions are the same as for 16. Rewrite as

f(x)=\dfrac4{2x+3}=\dfrac{\frac43}{1-\left(-\frac{2x}3\right)}

Now recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n

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f(x)=\dfrac43\displaystyle\sum_{n\ge0}\left(-\frac{2x}3\right)^n

Because this is a geometric sum, this converges when \left|-\dfrac{2x}3\right|, or |x|. This would be the interval of convergence.

Your hunch about checking the endpoints is correct. Checking is easy in this case, because at the endpoints (-3/2 and 3/2) the series obviously diverges.

- - -

16. This one is kind of tricky, and there's more than one way to do it. The standard method would be to take the antiderivative:

F(x)=\displaystyle\int f(x)\,\mathrm dx=\int\frac{\mathrm dx}{(1+x)^2}=-\frac1{1+x}+C

We also have

\displaystyle-\frac1{1+x}=-\frac1{1-(-x)}=-\sum_{n\ge0}(-x)^n\implies F(x)=C-1-\sum_{n\ge1}(-x)^n

and differentiating this gives

f(x)=-\displaystyle\sum_{n\ge1}n(-x)^{n-1}=-\sum_{n\ge0}(n+1)(-x)^n=\sum_{n\ge0}(n+1)(-1)^{n+1}x^n

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\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)(-1)^{n+2}x^{n+1}}{(n+1)(-1)^{n+1}x^n}\right|

The limit reduces to

\displaystyle|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|

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