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3 years ago

The proof ABC ≅ DCB that is shown.

Mathematics

2 answers:

miskamm [114]3 years ago

8 0

Solution:

Given: ∠A≅∠D and CD║AB

To Prove: Δ ABC ≅ ΔD CB

Proof:

1. ∠A=∠D→→(Given)

2. As, CD ║AB, ⇒ ∠DCB=∠ABC→→(when lines are parallel, Alternate interior angles are equal.)

3. Segment CB=Segment BC→→Reflection property

⇒Δ ABC ≅ ΔD CB →→(Angle Angle Side or AAS)

Schach [20]3 years ago

4 0

The <em><u>correct answer</u></em> is:

AAS

Explanation:

AAS stands for "angle-angle-side." This states that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle, then the triangles are congruent.

In these triangles, we have ∠CAB ≅ ∠CDB given to us to begin with. Throughout the proof, we find that ∠ABC ≅ ∠DCB. We also have that CB is congruent to itself. This is two angles and a side not included, so this is AAS.

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First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

Bill eats 3 slices of cornbread for breakfast that contain a total of 543 calories. How many

Verizon [17]

Each slice contains 181 calories

543/3=181

Brainliest?

8 0

2 years ago

HELP MEEEEE PLEASE I DONT UNDERSTAND THIS QUESTION?????

Vanyuwa [196]

Answer:

It's B

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help ASAP lol

MAXImum [283]

B and C are the correct answers

4 0

3 years ago

Math Help please 15 points

Law Incorporation [45]

1) You have vertical angles here. This means the big and small triangles are similar, meaning the ratio of corresponding sides of the two triangles are equal.

In this case, 32cm corresponds to 4cm and 40cm corresponds to x. To find x, set up the ratio as follows: 4/32 = x/40.

4/32 can be reduce to 1/8. Multiply each side by 40 to isolate x. you get 1/8 * 40

This yields x = 5 cm

2) set up a proportion, similar to the first question.

(x + 4)/8 = (2x + 1)/12
12(x + 4) = 8(2x + 1)
12x + 48 = 16x + 8
40 = 4x
x = 10

the length of BC is (x + 4) + (2x + 1) = 3x + 5

this means BC = 35

4 0

3 years ago