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3 years ago

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Peter uses unit cubes to build a figure in the shape of the letter X what is the fewest unit cubes that Peter can use to build t

he figure

1 answer:

lozanna [386]3 years ago

4 0

I am not sure I am right but I think it depends on the size of the X.

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

2 years ago

WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the

iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

$d=\sqrt{h^2+k^2}$

Distance of P from ( -3, 4 ) :

$d=\sqrt{(h+3)^2+(k-4)^2}$

Now , these distance are equal :

$h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2$

Solving above equation , we get :

$P=(\dfrac{16}{21},\dfrac{23}{7})$

Hence , this is the required solution.

6 0

3 years ago

If the coefficient of determination for a data set containing 24 points is 0.5,

aleksandrvk [35]

Answer:

False

Step-by-step explanation:

False - one has nothing to do with the other. None of the data points can lie on the regression line,

and you can have the coeff. of determination be 0.5.

7 0

3 years ago

If 6 1/4 of potatoes cost 40.75 how much would one pound cost

Darya [45]

Answer 6 dollars and 52 cents

Step-by-step explanation: 6 1/4 = (6.25)

40.75/6.25 = 6.52 per unit price

6 0

2 years ago

Find the value of x.<br> A<br> 28<br> 20<br> B<br> 9x – 8<br> С

natulia [17]

Answer:

The question is incorrect. please provide a diagram.

4 0

2 years ago