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maxonik [38]
3 years ago
13

Factor quadratic equation 6x.squared +41x + 63 = 0

Mathematics
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

x1= -9/2 ; x2= -7/3

Step-by-step explanation:

Rewrite to 6x^2 + 27x +14x +63 = 0

Factor and get 3x(2x + 9) + 7(2x + 9) = 0

Factor again and get (2x + 9)(3x + 7) = 0

Separate into possible cases -> 2x + 9 = 0; 3x + 7 = 0

The equation has two solutions: x1 = -9/2 and x2 = -7/3

Hope this helped!!

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Expand the given power using the Binomial Theorem. (10k – m)5
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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}

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3 years ago
Which data set does this stem-and-leaf plot represent?
Ludmilka [50]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Surface area = 2(6 × 4) + 2(6 × 4) + 2(4 × 4)

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Answer: -11

Step-by-step explanation:

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From there, plug in 3 for x and solve.

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Step-by-step explanation:

To get from F to C you do

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