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ddd [48]
3 years ago
15

What are the multiplication sentences for factors of 6?

Mathematics
1 answer:
mart [117]3 years ago
7 0

Answer:

keep doubling the answer by 6

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Hosea borrowed $20,000 to buy a new speed boat. The interest rate on his loan is 5%. If he plans to pay the loan back in three y
Anit [1.1K]

Answer:

a. Interest amount is $3,000.

b. Total cost is $23,000.

Step-by-step explanation:

a. How much interest will he pay

This can be calculated as follows:

Interest amount = Principal loan Amount * Interest rate * Number of years ........ (1)

Where;

Principal loan Amount = $20,000

Interest rate = 5%

Number of years = 3

Substituting the values into equation (1), we have:

Interest amount = $20,000 * 5% * 3 = $3,000

b. What will be the total cost that he will owe including interest?

Total cost = Principal loan Amount + Interest amount = $20,000 + $3,000 = $23,000

7 0
3 years ago
Which statement is true?
iren [92.7K]
It’s d hope this helps
4 0
3 years ago
The square root of a number is between 10 and 11. Which of these could be that number?
kaheart [24]
The answer would be 105, it's square root is approximately 10.2469 <span />
7 0
3 years ago
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A 6-pack of fruit juice costs $2.76. At that same rate, how much would a 15-pack cost
Scorpion4ik [409]
A 15 - pack would cost $6.90.

Explanation:
We know that a 6-pack of fruit juice would cost 2.76, so to find the cost of 1 -pack, you divide 2.76 by 6. That gives us $0.46 for 1 -pack. To find the cost of a 15-pack, you would multiply $0.46 by 15, which gives us $6.90. Therefore, a 15- pack would cost $6.90
3 0
3 years ago
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
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