All categories

2 years ago

You start at (2, 5). You move left 5 units and right 8 units. Where do you end?

Mathematics

2 answers:

lianna [129]2 years ago

7 0

Answer:

(5,5)

Step-by-step explanation:

left 5 is -5 and right 8 is +8.

-5+8=3

2+3=5

When talking about left and right, you change the x-coordinate, when up and down, y-coordinate.

Overall, (5,5)

Vitek1552 [10]2 years ago

4 0

Answer:

5,5

Step-by-step explanation:

since we are not changing the height of the point and just moving side to side, the y intercept does not change. Once you move 5 units to the left, your new point is -3,5 but then when you move 8 to the right it switches to positive again and the new point is 5,5

You might be interested in

Min-Jee is planting two rectangular flower beds. One bed is 2.78 feet by 4.81 feet, and the other bed is 2.78 feet by 5.61 feet.

shusha [124]

Answer: $(2.78+4.81)(2.78+5.61)$

Step-by-step explanation:

<h3> The complete exercise is attached.</h3><h3></h3>

The area of a rectangle can be calculated with this formula:

$A=lw$

Where "l" is the lenght and "w" is the width.

Then, you can notice that it can be obtained by multiplying the dimensions of the rectangle.

Knowing this, you can determine that the total area of the two flowers bed can be obtained by adding the products of their dimensions.

Since one of the rectangular flower bed is 2.78 feet by 4.81 feet and the other bed 2.78 feet by 5.61 feet, you can write the following expression to find the total area (in square feet)of the two beds:

$2.78*4.81 +2.78*5.61$

If you factor out 2.78:

$2.78(4.81 +5.61)$ or $(4.81 +5.61)2.78$

Therefore, the expression that does not represent the total area in square feet of the two beds, is:

$(2.78+4.81)(2.78+5.61)$

6 0

3 years ago

Read 2 more answers

Benchmark estimate of 8.01

Nana76 [90]

I think it would just be 8 or 8.00

5 0

3 years ago

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

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