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3 years ago

You have boxes that will hold 1 cubic foot, 3 cubic feet,

Mathematics

1 answer:

valkas [14]3 years ago

8 0

Answer:

13 boxes

Step-by-step explanation:

To calculate the minimum number of boxes that can be had with that amount of paper, they should be kept in the largest box, that is, the 27 cubic foot box, therefore, we divide:

257/27 = 9.51

Which means that we can save the paper in 9 boxes of 27 cubic feet, now we will save the rest in the next smaller box, the 9 cubic foot box.

First let's calculate how many leftover paper:

257 - 27 * 9 = 14

Therefore, we can only use 1 box of 9 cubic feet:

14 - 9 = 5

For the remaining 5 cubic feet use 1 box of 3 cubic feet and 2 boxes of 1 cubic foot.

Therefore, in total it would be:

9 of 27 ft ^ 3 + 1 of 9 ft ^ 3 + 1 of 3 ft ^ 3 + 2 of 1 ft ^ 3 = 13

a total of 13 boxes would be the minimum

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Brainliest question please help please help me and plz show your work

Alex777 [14]

Answer:

Step-by-step explanation:

since the radius is 12, the most the circle expands out towards is 15 for x and 15 for y. (-6,-5) easily fits inside it.

8 0

4 years ago

**Spam answers will not be tolerated**

Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

What multiplies to be -18 and adds to be -9?

Harrizon [31]

IMPORTANT:
If you're trying to factor a quadratic in Algebra I:
There are no two integers that can solve this problem!
Your quadratic is <em>prime</em>!

If you're trying to solve a quadratic (find x):
The factoring approach will not work for the same reasons listed above.
Try using splitting the middle or the quadratic formula instead.

Here's how you would solve it from a more advanced approach.
If you don't know what this stuff is, just ignore it.
ab = -18, a + b = -9
Find a in terms of b.
a = -9-b
Substitute this for a in the first equation.
(-9-b)b = -18
-9b-b² = -18
Multiply everything by -1 to get rid of all these negative signs.
b² + 9b = 18
Bring over that 18.
b² + 9b - 18 = 0
Apply the quadratic formula.
(a = 1, b = 9, c = -18)
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-9\pm\sqrt{9^2-4(1)(-18)}}{1\times2}=\frac{-9\pm\sqrt{81+72}}{2}=\frac{-9\pm\sqrt{153}}{2}=\boxed{\frac{-9\pm3\sqrt{17}}{2}}$
If you need to write two distinct numbers, just write out one with a + and one with a - in place of the plus-minus sign.

8 0

3 years ago

2n-3 sequence 1st term

Alexandra [31]

First term = 2(1) - 3 = 2 - 3 = -1

8 0

3 years ago

Please help pleasehelp

grandymaker [24]

Answer:

≥ and ≤ will both be solid dots, as it includes the number

x ≥ -7 is everything greater than -7; x ≤ 4 is everything less than 4.

to include both, you'd click -7, drag right, and stop at 4

7 0

2 years ago