All categories

3 years ago

Which expression represents the expanded version of the expression below? 4x(3x+7)

Mathematics

2 answers:

puteri [66]3 years ago

7 0

Answer:

X is just a variable.

Step-by-step explanation:

emmasim [6.3K]3 years ago

3 0

Umm i would say 12x^2 + 28x

You might be interested in

Y = x^2 - 4x + 4 Find the discriminant. Determine the number of real solutions.

weqwewe [10]

Answer:

The discriminant is 0

There is 1 real solution

Step-by-step explanation:

Use the discriminant formula, D = b² - 4ac

In the equation y = x² - 4x + 4, a is 1, b is -4, and c is 4.

Plug in these values into the formula:

D = b² - 4ac

D = (-4)² - 4(1)(4)

D = 16 - 4(4)

D = 16 - 16

D = 0

So, the discriminant is 0.

With a discriminant of zero, there is one real solution.

So, the number of real solutions is 1.

8 0

3 years ago

A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)

Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

$\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}$

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by x ²,

$\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}$

Then each remaining rational term converges to 0 as x gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

$\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}$

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than x - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

$\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}$

More precisely,

$\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0$

D) Looks like this limit should read

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}$

which is just another case of (A) and (B); the limit would be

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1$

That is,

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1$

However, in case you meant something else, such as

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}$

then the limit would be different:

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0$

since the degree of the denominator is larger.

One important detail glossed over here is that

$\sqrt{t^2} = |t|$

for all real t. But since t is approaching *positive* infinity, we have t > 0, for which |t | = t.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

$\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}$

Again,

$\sqrt{x^8} = |x^4|$

but x ⁴ is non-negative for real x.

F) Also somewhat ambiguous:

$\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3$

$\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty$

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large x, x ⁴ is much larger than x ², so effectively

$\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}$

6 0

3 years ago

Rachel pays $131.44 for a desk, including tax. the tax is 31 cents less than 1/16 of the desk's price. what is the price of the

Paha777 [63]

$123.53 because divide 132.44 by 16 and subtract .31 is the tax. Subtract that number from 131.44 is the cost of the desk before tax

5 0

3 years ago

An airline company advertises that 100% of their flights are on time after checking 5 flights from yesterday and finding that th

just olya [345]

a) The population is all of the flights belonging to the particular airline. It's not all flights in general because the airline does not have control over a rival company's flight schedule.

------------------

b) The sample is the 5 airlines the company selected. Ideally the sample should represent the population as much as possible. The larger the sample, the more representative the sample is.

------------------

c) No, the sample is likely not representative because one day of flight does not represent the entire lifetime of all flights done so far for that company. This particular day could have been a very good day with good weather, which may explain why the 5 flights were all on time.

------------------

d) It would be better for the company to select the days at random and sample all of the flights for those particular days chosen. This is a cluster sample. Each cluster is a day. Also, I think more flights should be sampled. Five flights does not seem like enough.

8 0

3 years ago

Which of the following is closest to the circumference of a circle whose radius is 21 m

Sloan [31]

Here Is your answer!!!!

7 0

3 years ago