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4 years ago

15

Mr. Case is placing boxes along one of the walls in his attic. If the wall is 11 1⁄3 feet long and each box is 2 1⁄2 feet long,

how many boxes can he fit along the attic wall? boxes (Hint: Only a whole number of boxes can be placed along the wall.)

1 answer:

Alika [10]4 years ago

7 0

4.532 boxes

I am writing this because I need 20 characters

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What is the solution to the equation 13 3/4 + x equals 7 1 / 4

Olegator [25]

Answer:

x=-6.5

Step-by-step explanation:

$7\frac{1}{4} - 13\frac{3}{4}=-6\frac{1}{2}=-6.5$

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4 years ago

Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0

marin [14]

If you do in fact mean $f(1)=f(6)=0$ (as opposed to one of these being the derivative of $f$ at some point), then integrating twice gives

$f''(x) = -\dfrac1{x^2}$

$f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1$

$f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2$

From the initial conditions, we find

$f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0$

$f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)$

Eliminating $C_2$ , we get

$(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))$

$-5C_1 = \ln(6)$

$C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)$

Then

$\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}$

3 0

2 years ago

Simone paid $5.25 for 3 pounds of apples. Which equation represents the cost, in dollars, C, of buying a pounds of apples?

Shalnov [3]

I believe it is the first once because if a means the amount of apples and $1.75 is the amount paid for 1 apple, this answer would be correct.

5 0

3 years ago

510 exercise books were shared among schools X, Y and Z. such that Z had 3/2 times as much as Y; and Y had 3 times as much as X.

ycow [4]

Overall, the goal is to find the value of x and then use this and the given total to find the percentage of x. First, to do this, we must set up our equations based on the given information.

x + y + z = 510

z = (3/2)y

y = 3x

Now, we can begin to solve with substitution. The ultimate goal is to get the first equation to be regarding one variable and them work backwards using that found value.
Substitute the 3rd equation into the second.

z = (3/2)(3x) ----> Simply
z = (9/2)x

Now, when we plug in z and y into the first equation, only x will be present.

x + (3x) + (9/2)x = 510 ---> Combine like terms.
4x + (9/2)x = 510 ---> Get a common denominator.

(8/2)x + (9/2)x = 510

(17/2)x = 510 -----> Multiply both sides by 2/17

Now, if you understand what you are looking at, you know that you just found your answer which is 2/17 because you are about to multiply that by the total.

2/17 = .12 × 100 = 12%

If you did not know that, you can continue solving and then find the answer using part over whole or the total books of school x divided by the total books from all schools. You will get the same answer!

x = (2/17)×510
x = 60

60/510 = .12 × 100 = 12%

I hope this helps!

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4 years ago

Which of the following values in the set below will make the equation 5x + 6 = 6 true?

Rzqust [24]

$5x + 6 = 6 \\ \\ 5x=6-6 \\ \\ 5x=0 \\ \\ x=0 \\ \\ \text{The answer is x=0}$

6 0

4 years ago