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3 years ago

. Simplify the sum. (2u3 + 6u2 + 2) + (7u3 – 7u + 4)

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

Answer:

9u^3 + 6u^2 - 7u + 6

Step-by-step explanation:

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The point slope equation of a line is?

snow_tiger [21]

Answer:

$\text{The point-slope equation of a line is:}$

$C.\ y-y_0=m(x-x_0)$

$m-\text{slope}\\\\(x_0,\ y_0)-\text{point on a line}$

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3 years ago

Probability PLZ HELP

Y_Kistochka [10]

Answer:

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Step-by-step explanation:

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3 years ago

Suppose I make several time measurements, and compute an average time of 3.4 s, with a standard deviation of 0.8 s and an SDOM o

7nadin3 [17]

Answer:

The result will be smaller than 2.6s, 16% of the time.

Step-by-step explanation:

This is a normal distribution problem

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ

x = the random value = 2.6 s

xbar = mean = the average = 3.4 s

σ = standard deviation = 0.8 s

z = (2.6 - 3.4)/0.8 = - 1

P(x < 2.6) = P(z < - 1) = 1 - P(z ≥ - 1) = 1 - P(z ≤ 1) = 1 - 0.841 = 0.159 = 16%

8 0

3 years ago

Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ModifyingBelow lim W

Sergeu [11.5K]

Answer:

The value of given limit problem is 0.

Step-by-step explanation:

The given limit problem is

$lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}$

We need to find the value of given limit problem.

Divide the numerator and denominator by the leading term of the denominator, i.e., $x^4$

$lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}$

$lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}$

Apply limit.

$\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}$

We know that $\frac{1}{\infty}=0$ .

$\dfrac{0+0-0}{6-0-0}$

$\dfrac{0}{6}$

$0$

Hence, the value of given limit is 0.

8 0

3 years ago

Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic a

Sliva [168]

Answer:

0.2889 g brominated product

64.6 %

Step-by-step explanation:

This is a bromination chemical reaction of an alkene and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.

The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.

trans-cinnamic acid + pyridinium tribromide ⇒ 2,3-dibromo-3-

phenylpropanoic acid

Molar weight trans-cinnamic acid = 148.16 g/mol

mass trans-cinnamic acid = 139.0 mg x 1g/1000 mg = 0.139 g

# mol trans-cinnamic acid = 0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol

Since our reaction is 1 mol trans-cinnamic acid produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:

1 mol 2,3-dibromo-3-phenylpropanoic acid / trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid

= 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid

In grams the the theoretical yield is:

molar mass 2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol

The theoretical mass 2,3-dibromo-3-phenylpropanoic acid:

= 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol

= 0.2889 g

% yield = mass experimental/mass theoretical

= 0.1866 g / 0.2889 g x 100 = 64.6 %

4 0

3 years ago