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4 years ago

Two competing gyms each offer childcare while parents work out

Mathematics

1 answer:

KIM [24]4 years ago

8 0

I'm sorry, what is the question here?

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What number is 27% of 52? Round your answer to the nearest hundredth if 1 point necessary

uranmaximum [27]

Answer:

14.01

Hope this helps!!!!

Have a great day!!!!! :)

8 0

3 years ago

Read 2 more answers

I need help, if is right I mark you as brainliest.

borishaifa [10]

Answer:

D. He distributed -5 incorrectly.

Step-by-step explanation:

Last option is the correct answer.

5 0

3 years ago

Read 2 more answers

Determine the value Y, if X is -2 Y=X-12

8090 [49]

10 because -2 minus 12 is 10

3 0

2 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

Mia bought 10 boxes of brownies at the store for a total of $64. The tax on the purchase was $5. What was the cost of each box?

Pavel [41]

Let r represent cost of each box.

We have been given that Mia bought 10 boxes of brownies. So cost of 10 boxes would be $10r$ .

We have been given that Mia bought 10 boxes of brownies at the store for a total of $64. The tax on the purchase was $5. So the total cost of 10 boxes would be cost of 10 boxes plus tax.

We can represent this information in an equation as:

$10r=64+5$

We can subtract 64 from both sides as:

$10r-64=64-64+5$

$10r-64=5$

Therefore, our required equation is $10r-64=5$ and option 'c' is the correct choice.

7 0

3 years ago