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2 years ago

20. Carmen can buy bottles of paint for $2.00 each and boxes of colored pencils for $3.50 each. She can spend no more than

Mathematics

1 answer:

KIM [24]2 years ago

6 0

Answer:

(5, 8), (7, 1), (15, 3)

Step-by-step explanation:

a) The variables are defined in the problem statement. The total expenditure must satisfy ...

2.00x +3.50y ≤ 42.00

b) Possible solutions include ...

(x, y) ∈ {(5, 8), (7, 1), (15, 3)}

These were chosen by looking at the attached graph of the equation.

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The equation 12.5(x + 20) = 460 describes Steve's exercise plan for the

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B 38.4 miles because it’s obvious I took the test trust me

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3 years ago

Use the expression 4z + 8 + 3x + b to answer the following

nikdorinn [45]

Answer:

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4z, 8, 3x, and b

part b:

part c:

4 and 3

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2 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

Steven buys 24 bags of flour. Each bag holds 25 pounds of flour. How much flour did Steven buy?

frez [133]

If he has 24 bags and each is 25 pounds, you need to multiply:

24x25=600
Steven bought 600 pounds of flour.

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3 years ago

Round this number to the nearest tenth.<br><br> 52.85

k0ka [10]

Answer:

52.9

numbers 0-4 the number stays the same

numbers 5-9 the number goes up one

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3 years ago