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2 years ago

Tell whether ΔABC and ΔDCB can be proven congruent.

Mathematics

1 answer:

Vikentia [17]2 years ago

4 0

Answer:

D. No, there isn’t enough information because only two pairs of corresponding sides can't be used to prove that two triangles are congruent.

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Which of the following is a solution to the inequality below? 12 - x/11 > 8

n200080 [17]

Answer:

12 - x/11 > 8

12 -8 - x/11 > 8 -8

4 -x/11 > 0

4 -x/11 + x/11> 0 + x/11

4 > x/11

44 > x

x is less than 44

Step-by-step explanation:

7 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

2 years ago

Third term of (x + 3) 12

dem82 [27]

12 I’m pretty sure because you have x then 3 then 12

5 0

2 years ago

True of false !!! The triangle shown below must be congruent

user100 [1]

False.

Right triangles do not have to be congruent.

6 0

2 years ago

Read 2 more answers

Here is a diagram and its corresponding equation. Find the solution to the equation.

Katen [24]

4(x + 7) = 38

Since there is no operation sign between the 4 and parenthesis, it makes it an automatic multiplication problem.

You would use distributive property to multiply each which should leave your equation looking like this afterwards:

4x + 28 = 38
(4 times x and 4 times 7)

You would then solve it like a two step equation,

1. +28 - 28 = 0
2. 38 - 28 = 10

4x = 10

10 divided by 4 = 2.5

x = 2.5

We can recheck our work by substituting in the value of x.

4(2.5 + 7) = 38

10 + 28 = 38

When we substituted in the value, it gave us the correct inequality.

Therefore,
x = 2.5

5 0

2 years ago

Read 2 more answers