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Alekssandra [29.7K]
3 years ago
10

Factorise 4a²_4ab+b²​

Mathematics
1 answer:
Natalija [7]3 years ago
7 0

Answer:

(2a -b)^2

Step-by-step explanation:

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Solve fast answer gets brain and its 30 points ty
Vlad1618 [11]

Answer:

74

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The probability of getting a head when a biased coin is tossed is . The coin is tossed three times. Find the following probabili
S_A_V [24]

Answer:

a) P(three tails) = \frac{1}{8}

b) P(two tails followed by one head) = \frac{1}{8}

Step-by-step explanation:

coin is tossed three times so, outcome is

HHH, TTT, HTH, THH, HHT, THT, TTH, HTT.

Hence sample space will be

S = {HHH, TTT, HTH, THH, HHT, THT, TTH, HTT}

chance of getting three tails is '1' i.e. TTT

hence,

P(three tails) = \frac{1}{8}

two tails followed by one head is TTH

so, chance of getting that outcome is also '1'

hence,

P(two tails followed by one head) = \frac{1}{8}

6 0
3 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
if A is the number of degrees and B is the number of grades of any angle ,prove that : B=A+A/9 [FULL PROCESS REQUIRED]​
Gre4nikov [31]
We know
90
∘
=
1
right angle
=
100
g
So
A
∘
=
(
100
90
⋅
A
)
g
Hence by the given condition
B
=
(
100
90
⋅
A
)
=
10
9
A
=
A
+
(
A
9
)
Answer

3 0
3 years ago
Find a16 of the sequence 1,6,11,16
babymother [125]

\bf 1~~,~~\stackrel{1+5}{6}~~,~~\stackrel{6+5}{11}~~,~~\stackrel{11+5}{16}...\qquad \qquad \stackrel{\textit{common difference}}{d=5} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=1\\ d=5\\ n=16 \end{cases} \\\\\\ a_{16}=1+(16-1)5\implies a_{16}=1+(15)5\implies a_{16}=76

6 0
3 years ago
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