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3 years ago

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HELP ASAP PLEASE!1!! Determine the factors of a2b + 2a2 + 3b + 6. (b + 3)(a2 + 2) (b + 2)(a2 + 3) (ab + 2)(a + 3) (ab + 3)(a + 2

)

1 answer:

andrey2020 [161]3 years ago

7 0

The answer is (b+2)(a²+3).
Hope this helps!

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Solve:<br> 10c + 20 = 40<br> c =

Dvinal [7]

C= 2

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8 0

3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

Zucchini weights are approximately normally distributed with mean 0.8 pound and standard deviation 0.25 pound. Which of the foll

Charra [1.4K]

Answer:

0.81859

Step-by-step explanation:

We solve using z score method

The formula for calculating a z-score is is z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Mean 0.8 pound

Standard deviation 0.25 pound

For x = 0.55 pound

z = 0.55 - 0.8/0.25

= -1

Probability -value from Z-Table:

P(x = 0.55) = 0.15866

For x = 1.3 pounds

Z = 1.3 = 0.8/0.25

= 2

Probability value from Z-Table:

P(x = 1.3) = 0.97725

The probability that a randomly selected zucchini will weigh between 0.55 pound and 1.3 pounds is

P(x = 1.3) - P(x = 0.8)

0.97725 - 0.15866

0.81859.

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3 years ago

1. What is the simple interest on $4,500 for 2 and a half years at 4% per year?

allochka39001 [22]

Answer: 1st is 180$ sorry if I’m wrong

Step-by-step explanation:

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3 years ago

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seropon [69]

C) 5.5 thats what I got.

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3 years ago

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