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2 years ago

Can someone help me in this? a^2-ab-b-1

Mathematics

1 answer:

zheka24 [161]2 years ago

7 0

Answer:

(a + 1) (a − b − 1)

Step-by-step explanation:

I assume you're trying to factor the expression? If so, use quadratic formula to find the roots.

a = [ -(-b) ± √((-b)² − 4(1)(-b−1)) ] / 2(1)

a = [ b ± √(b² + 4b + 4) ] / 2

a = [ b ± (b + 2) ] / 2

a = -1 or b + 1

(a − (-1)) (a − (b + 1))

(a + 1) (a − b − 1)

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5. Landon bought a new car for $16,000 and it depreciates 4.5% every year. Write a function that describes

snow_tiger [21]

Answer:

<h3>The value C(t) of the car after 5 years is $12709.</h3>

Step-by-step explanation:

Given that Landon bought a new car for $16,000 and it depreciates 4.5% every year.

<h3>To find the value C(t) of the car after 5 years:</h3>

Initial value $C(0)= $16,000$

Depreciation rate is $r=\frac{4.5}{100}$

Period , t=5 years

$C(t)=C(0)(1-r)^t$

Substitute the values we get

$C(5)=16000(1-0.045)^5$

$=16000(0.955)^5$

$=16000(0.7943)$

∴ $C(5)=12708.8$

<h3>The value C(t) of the car after 5 years is $ 12709</h3>

3 0

2 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

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2 years ago

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gogolik [260]

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3 years ago

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Vedmedyk [2.9K]

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Step-by-step explanation:

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2 years ago

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Vanyuwa [196]

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Hope this helps!

6 0

2 years ago