Fill in the missing number.<br> 6 x 4 =(<br> x 4) + (3 x 4)

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$

6 0

3 years ago

Simplify the expressions and find the answer.

cupoosta [38]

3- 5/9= 2.45
3- 23/9= .45

2.45/.45= 5.44

5.44^3/2= 81

6 0

3 years ago

Find each percent decrease of 80 to 64

Anastasy [175]

80-64=24
24 is 30% of 80 so a decrease of 30%

4 0

3 years ago

Easy math, but not sure what I did wrong.

Ierofanga [76]

Answer:

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year. So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3. However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6. Same problem as before.

If a₁ = 3, then a₅ = 9. This is a possible solution.

If a₁ = 4, then a₅ = 12. If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year. So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0

3 years ago

Trisha needs to make at least 50 gift bags for an event. Each gift bag will contain at least 1 thumb drive or 1 key chain. She w

Anon25 [30]

There will be 50 gift bags.

The first set of 25 bags
----------------------------------
Because there are 25 thumb drives, the first 25 bags will each have 1 thumb drive.
Also, each bag will have 5 times as many key chains as a thumb drive. Therefore each of the first 25 bags will have
1 thumb drive, 5 key chains.

After the first 25 bags, we have used
25 thumb drives, 5*25 = 125 key chains.

The second set of 25 bags
----------------------------------------
We have a total of 200 key chains, so we have 200-125 = 75 key chains left.
Distribute them equally among the remaining 25 bags, so each bag has
75/25 = 3 key chains.

Answer:
25 bags, each with 1 thumb drive and 5 key chains.
Another 25 bags, each with 3 key chains.

4 0

2 years ago

Read 2 more answers

Fill in the missing number. 6 x 4 =( x 4) + (3 x 4)