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3 years ago

If a△ b=a2+ab−b2, then find (3 △ 2) △ 13. WILL MARK BRAINLIEST PLS HELP!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Olenka [21]3 years ago

7 0

Answer:

The answer is

Step-by-step explanation:

a ∆ b = a² + ab + b²

(3 ∆ 2)

a = 3 b = 2

(3 ∆ 2) = 3² +3(2) + 2²

= 9 + 5 + 4

= 9 + 9

= 18

Hope this helps you

Andreas93 [3]3 years ago

3 0

<h3>Answer: 95</h3>

=====================================================

Work Shown:

First we compute 3 △ 2

a △ b = a^2 + ab - b^2

3 △ 2 = 3^2 + 3*2 - 2^2 ... replace 'a' with 3, replace b with 2

3 △ 2 = 9 + 6 - 4

3 △ 2 = 15 - 4

3 △ 2 = 11

Replace the ( 3 △ 2 ) in the original expression with 11.

We now want to find 11 △ 13

a △ b = a^2 + ab - b^2

11 △ 13 = 11^2 + 11*13 - 13^2

11 △ 13 = 121 + 143 - 169

11 △ 13 = 264 - 169

11 △ 13 = 95

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The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

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$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

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The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :