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3 years ago

A2 + 2ab + b2-9 Pls help

Mathematics

2 answers:

Anarel [89]3 years ago

8 0

Answer:

(a + b + 3 ) ( a + b − 3 )

Step-by-step explanation:

valentinak56 [21]3 years ago

4 0

Answer:

${\huge{\mathfrak{\underbrace{\pink{Answer}}}}}$

I hope this helps you.....

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My sister needs help... blahhh!!! please answer 8,9, and 10 please!!!! Thank you!

castortr0y [4]

8:1 -4/15 9:6 1/2 10:can't figure it out!

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3 years ago

Consider the equation below and determine the value of k

Ksenya-84 [330]

Answer:

k=9

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3 years ago

Find the volume of the cylinder. Round your answer to the nearest tenth.

ludmilkaskok [199]

Answer:

1526.8 ft ^3

Step-by-step explanation:

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3 years ago

Read 2 more answers

The net of an isosceles triangular prism is shown. What is the surface area, in square units, of the triangular prism?

neonofarm [45]

Area of three rectangles

3LB
3(8)(6)
24(6)
144units²

Area of two triangles

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Total

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5 0

2 years ago

Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-

Charra [1.4K]

Answer:

The statement is true is for any $n\in \mathbb{N}$ .

Step-by-step explanation:

First, we check the identity for $n = 1$ :

$(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}$

$1 = \frac{1\cdot 1\cdot 3}{3}$

$1 = 1$

The statement is true for $n = 1$ .

Then, we have to check that identity is true for $n = k+1$ , under the assumption that $n = k$ is true:

$(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)$

$2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)$

$(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)$

Therefore, the statement is true for any $n\in \mathbb{N}$ .

4 0

3 years ago