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3 years ago

HOW WOULD THE NUMBER LINE MODEL BE DIFFERENT IF YOU WANTED TO FIND (-5) + (-6)=

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

Step-by-step explanation:

(-5)+(-6) is the same thing As -5-6 so you can move 6 “jumps” backward starting -5

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Calculate 0.82 + 0.35 ?

natima [27]

ANSWER

EXPLANATION

We want to add the two numbers:

0.82 + 0.35

Using the long addition method, we have:

0 . 8 2

+ 0 . 3 5

Start on the right hand side:

4 0

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

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

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

8 0

3 years ago

3+x=3 whats the answer its 100 points

schepotkina [342]

X is 0 so that means you answer would be 3+0=3

5 0

3 years ago

What is the length of EF?

ANEK [815]

The slope is 5/4.

5²+4²=41
√41=6.4
EF=6.4 units.

3 0

3 years ago

Read 2 more answers

Which statement regarding the graph is true? Which of the following statements is true?

mylen [45]

Which statement regarding the graph is true? Which of the following statements is true?

Question:

The store sold 4,000 fewer units in November than in August.

Answer:

FALSE.

Explanation:

Number of units sold in November = 3000

Number of units sold in August = 6000

difference in both = 6000-3000= 3000

which not same as 4000 hence wrong

Question:

More customers prefer chocolate than strawberry and vanilla combined.

Answer:

FALSE.

Explanation:

value for Chocolate is 38% which is clearly less than combined value of strawberry (30%) and vanilla (22%). which is (30+22=52%)

Question:

Yogurt sales had a decreasing trend during the last four months of 2011.

Answer:

TRUE

Explanation:

Graph is going down ward for the last four months (sept to Dec) which means decreasing sales.

Question:

The median number of ounces sold from 3 to 4 p.m. was more than the median number sold from 7 to 8 p.m

Answer:

FALSE.

Explanation:

It is not true for all Stem values.

Like when stem =1, then (3 to 4 p.m.) has median =0

but for (7 to 8 p.m) has median 6 which is more than 3 to 4 pm

5 0

4 years ago