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3 years ago

What is the expanded notation for 15,729

Mathematics

1 answer:

svp [43]3 years ago

8 0

10000 + 5000 + 700 + 20 + 9

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What is 167,003 X 409,834= <br><br>What is the answer?

olga_2 [115]

Answer:

68,443,507,502

Step-by-step explanation:

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2 years ago

Jdibiobaaaaaaaaaaaaaaaaaaaaargdiofhwberepjdefihgbaudfnsklmnfarbgtishdfz kbhyguno

Scorpion4ik [409]

Nnnnnnnnn. sksjsjen

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3 years ago

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Use an inequality to describe the interval of real numbers <br><br>(-3,8]

Naya [18.7K]

38 i think it means infinite

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3 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago

Solve the quadratic equation by completing the square x^2-2ax+b=0

luda_lava [24]

Let a, ß be the roots of x² + 2ax + b =0.......[1]

⇒a + ß = -2a and aß = b

By hypothesis

la-ß | ≤ 2m

⇒ (a-ß)² ≤ 4m²

⇒ (a+ß)²-4aß < 4m²

⇒ 4a²-4b ≤ 4m²

⇒a² - b ≤ m²......(2)

And discriminant of (1) is >0

⇒4a²-4b> 0

⇒ b <a²...... (3)

From (2) and (3)

a² > b≥ (a² - m²)

b = [a² - m², a²)

8 0

3 years ago