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kolbaska11 [484]

2 years ago

10

I purchased my very first chess set in June 1987. Since then, I have been steadily building my collection. In March 2009, I purc

hased my 14th set. Write a linear equation

1 answer:

steposvetlana [31]2 years ago

3 0

Answer:Ggfghj

Step-by-step explanation:

Ffhhjjj

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Whats the answer for (5x + 8)(4x² + 6x - 8)

mestny [16]

$(5x + 8)(4x^2 + 6x - 8)\\ \\ 5x(4x^2 + 6x - 8) + 8(4x^2 + 6x - 8) \\ \\ 20x^3 + 30x^2 - 40x + 8(4x^2 + 6x - 8) \\ \\ 20x^3 + 30x^2 - 40x + 32x^2 + 48x - 64 \\ \\ 20x^3 + (30x^2 + 32x^2) + (-40x + 48x) - 64 \\ \\ 20x^3 + 62x^2 + 8x - 64 \\ \\$

The final result is: 20x^3 + 62x^2 + 8x - 64.

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3 years ago

FREE 100 POINTS JUST SAY “SHAMALAMAMOOMOO”

Arisa [49]

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2 years ago

Read 2 more answers

If x^2+y^2=1, what is the largest possible value of |x|+|y|?

Marianna [84]

If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).

Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).

If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.

• Case 1: suppose <em>x</em> < 0. Then

<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2

• Case 2: suppose <em>x</em> ≥ 0. Then

<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2

In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.

6 0

2 years ago

Quality control is an important issue at ACME Company which manufacturers light bulbs. In order to conduct testing of the life h

mrs_skeptik [129]

Answer: 3424.28

Step-by-step explanation:

Given data : 378, 361, 350, 375, 200, 391, 375, 368, 321

Number of data values = 9

Mean : $\overline{x}=\dfrac{\sum^n_{i=1} x_i}{n}$

$\\\\=\dfrac{378+361+350+375+200+391+375+368+321}{9}\\\\=\dfrac{3119}{9}\approx346.56$

Variance = $\dfrac{\sum^n_{i=1} (x_i-\overline{x})^2}{n-1}$

$\sum^n_{i=1} (x_i-\overline{x})^2 = (31.44)^2+(14.44)^2+(3.44)^2+(28.44)^2+(-146.56)^2+(44.44)^2+(28.44)^2+(21.44)^2+(-25.56)^2\\\\=27394.2224$

Variance = $\dfrac{27394.2224}{8}=3424.2775\approx3424.28$

4 0

3 years ago

spayn [35]

Answer:81

Step-by-step explanation:

3 0

2 years ago