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2 years ago

What is the fourth term in the binomial expansion (a+b)^6)

Mathematics

1 answer:

Dafna11 [192]2 years ago

7 0

Answer:

$20a^3b^3$

Step-by-step explanation:

Binomial Series

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

Factorial is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$

$\implies \left(\dfrac{120}{6}\right)a^{3}b^3$

$\implies 20a^3b^3$

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Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

2 years ago

The common difference in an arithmetic sequence is –2 and the first term is 47. What is the 29th term?

BlackZzzverrR [31]

An arithmetic sequence is an ordered list of numbers where the next number is found by adding on to the last number (ex: 2,5,8,11... is a sequence where 3 is added on to find the next number).

The equation for an arithmetic sequence is
$A_{n}=A_{1}+(n-1)d$

$A_{n}$ is the "n-th" number in the sequence (ex: is the first term in the sequence)
d is the number you add (common difference) to find the next number
The first number in the sequence is 47 so $A_{1}=47$
d=-2 because the question gives you that

$A_{n}=A_{1}+(n-1)d$
 $A_{29}=47+(29-1)(-2)$
 $A_{29}=47+(28)(-2)$
 $A_{29}=47+-56$
 $A_{29}=-9$

The answer is A. -9.

 

The answer is C. 158

For the second one, it gives you $A_{1}=4$ and $A_{}=18$
You can use this to find d

 $A_{n}=A_{1}+(n-1)d$
 $A_{3}=4+(3-1)d$
 $18=4+(3-1)d$
 $18=4+2d$
 $14=2d$
 $7=d$

Now you can just solve using the equation normally.
 $A_{n}=A_{1}+(n-1)d$
 $A_{23}=4+(23-1)(7)$
 $A_{23}=4+(22)(7)$
 $A_{23}=4+154$
 $A_{23}=158$

The answer is C. 158

8 0

4 years ago

Read 2 more answers

Amari claims that a rectangle is sometimes a parallelogram but a parallelogram is always a

Maslowich

Answer:

The statement is false.

Step-by-step explanation:

A parallelogram is a figure of four sides, such that opposite sides are parallel

A rectangle is a four-sided figure such that all internal angles are 90°

Here, the statement is:

"A rectangle is sometimes a parallelogram but a parallelogram is always a

rectangle."

Here if we found a parallelogram that is not a rectangle, then that is enough to prove that the statement is false.

The counterexample is a rhombus, which is a parallelogram that has two internal angles smaller than 90° and two internal angles larger than 90°, then this parallelogram is not a rectangle, then the statement is false.

The correct statement would be:

"A parallelogram is sometimes a rectangle, but a rectangle is always a parallelogram"

6 0

3 years ago

And also this question as well

gtnhenbr [62]

Answer:

It's the second one I believe, because it's not a straight line, and nearly all functions have curved lines

Step-by-step explanation:

4 0

3 years ago

Andrews [41]

Answer:

The area of ΔABC= 6.667 square units

Step-by-step explanation:

ΔABC is similar to ΔMNO.

The scale factor from ΔMNO to ΔABC is 3∕2

the area of ΔMNO is 10 square units,

The area of ΔABC/the area of ΔMNO

= 2/3

The area of ΔABC/10= 2/3

The area of ΔABC= 2/3 * 10

The area of ΔABC= 20/3

The area of ΔABC= 6 2/3

The area of ΔABC= 6.667 square units

6 0

3 years ago