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4 years ago

(-6) ÷ (-7) in fraction form

Mathematics

1 answer:

pashok25 [27]4 years ago

8 0

Answer:

$\frac{6}{7}$

Step-by-step explanation:

The division symbol is also the middle line in a fraction. Since both are negative it results in a positive fraction.

Hope that helped and have a great day!

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Ms.Waddell put a 12-inch tall bucket under a leak in her sink. The bucket fills at a constant rate of 1/2 inch in every 1/6 of a

Angelina_Jolie [31]

Answer:

4 hours

Step-by-step explanation:

To solve this question you will simply make a ratio type equation. You know that every 1/6 of an hour, or every 10 minutes, the bucket will fill 1/2 an inch. The bucket is 12 inches tall.

$\frac{10 min}{1/2 inch} = \frac{x min}{12 inches}$

Cross multiply

$120=\frac{1}{2} x$

$x=240 min$

Convert minutes to inches

$\frac{240 min}{1} * \frac{1 hr}{60 min} = 4 hr$

Final answer: 4 hours

Hope this helps!

8 0

4 years ago

Shoutout to my sister amckeon88

kotykmax [81]

Periodt Pooh. For brainly??

4 0

3 years ago

Which expression gives the x-coordinate of the solution of the system 3x - y + z = 9?

Zolol [24]

×=y/3-z/3+3 thats the x value in this equation

8 0

3 years ago

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

3 years ago

Use the drop-down menus to complete each equation so the statement about its solution is true.

disa [49]

Answer:

Step-by-step explanation:

So,it would be easier to simplify the right equation, so when I say "the right equation", I mean the one I will make right now. It is 8x+9. So the first problem is 8x+9=x+?. Well, if there are no solutions, the slopes have to be the same, and the y-int doesn't matter. So the answer would be + 7x because that will make the x an 8x. If there is one solution, then the slope has to be different. So literally anything but 7x will work. Even constants. If there are infinitely many solutions, then the equations have to be the same. Meaning, you would have to add 7x and 9 to make the left equation the same as the right.

6 0

3 years ago

(-6) ÷ (-7) in fraction form​

(-6) ÷ (-7) in fraction form