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insens350 [35]
3 years ago
13

Evaluate the expression. 3 + (50- 5^2)

Mathematics
1 answer:
Mashutka [201]3 years ago
7 0
The answer is = 28 …………….
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Expand the given power using the Binomial Theorem. (10k – m)5
agasfer [191]

Answer:

(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}

∵ The binomial is (10k-m)^{5}

∴ a = 10k , b = -m and n = 5

∴ (10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ (10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})

∴ (10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

5 0
3 years ago
What is the expression in factored form?
san4es73 [151]

Answer:  (4x + 3) (x + 2)

Step-by-step explanation:

We can rewrite 4x² + 11x + 6 as 4x² + 8x + 3x + 6

Since 4x² + 11x + 6  = 4x² + 8x + 3x + 6

                                = (4x + 3) (x + 2)

6 0
3 years ago
Read 2 more answers
Find the midpoint of the line segment with the given endpoints. (1, 3), (3, 1)
Lilit [14]

Answer:

(2,2)

Step-by-step explanation:

6 0
3 years ago
What is the answer to this question
Gre4nikov [31]
It’s 4.08x10 to the 6 power. Which is answer A
6 0
3 years ago
If p q r is prime numbers such that pq+r=73, what is the least possible value of p+q+r
Jobisdone [24]
The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number. 

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58
6 0
3 years ago
Read 2 more answers
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