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3 years ago

Please help simplify 5y + 3 + 8y - 2

Mathematics

2 answers:

Vlad1618 [11]3 years ago

5 0

Answer:

13y+1

Step-by-step explanation:

You have to combine the like terms. And 5y+8y=13y. So the first number will be 13y, and you will keep the sign. So that would make it 13y+x. Then you will add 3+-2, which equals 1. So the answer is 13y+1

expeople1 [14]3 years ago

4 0

Answer:

13y + 1

Step-by-step explanation:

8y + 5y = 13y

3 - 2 = 1

13y + 1

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Twice the first number is eleven more than the sum of the other two numbers. The sum of twice the first number and three times t

lidiya [134]

Well first you have to make variables for each number.
1st - x
2nd - y
3rd - z

first sentence says twice first number so 2x. is means equals so =. eleven more than sum of the other two numbers. sum of other two numbers is (y + z) and eleven more than that is (y + z) + 11. So so this sentence says:
2x = (y + z) + 11

second sentence says sum of twice the first and three times third. twice first is 2x and three times third is 3z. their sum would be (2x + 3z). It it says is one more than second number. So = (y + 1). So so this means:
(2x + 3z) = (y + 1)

third sentence says second number (y) is is equal to sum of first and third (x + z). so;
y = (x + z)

now for the work.

We we can easily solve for a by subtracting the first 2 equations. the way to subtract 2 equations is by subtracting one left side of equal sign from the other equations left side and then doing the same with the right side.

So we will subtract the 1st and second equations we made.
So so left sides would be
2x - (2x + 3z)
2x - 2x - 3z
-3z

right side would be:
y + z + 11 - (y + 1)
y + z + 11 - y - 1
z + 10

now put the sides with an equal sign
-3z = z + 10
-4z = 10
z = -2.5

now we can plug in z into the equations and subtract second and third equations. But but we will subtract opposite sides. So so left minus right and right mi is left after we plug in z:
(2x + 3 (-2.5)) - (x + (-2.5))
2x - 7.5 - x + 2.5
x - 5

other one would be
(y + 1) - y
1

So so put an equal sign and get:
x - 5 = 1
x = 6

now plug x and x into 3rd equation
y = x + z
y = 6 + (-2.5)
y = 3.5

now we have values. You can check answer by plugging values into other 2 equations

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

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