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lyudmila [28]
3 years ago
13

If a box has a surface area of 680 and the length and width are 10. What is the height?

Mathematics
1 answer:
Molodets [167]3 years ago
8 0

10x10=100

680/100=6.8

10x10x6.8=680

the answer is 6.8

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b. 3 ice cream cones cost $8.25. At this rate, how much do 2 ice cream cones cost? Explain how you know.
Contact [7]
8.25/3 = $2.75 per ice cream
2.75 * 2 = $5.50
It would cost $5.50 for 2 ice cream
6 0
2 years ago
A cylindrical 2014-t6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis; results of these tes
Pani-rosa [81]

Complete part of Question:

Assume a factor of safety of 3.0

The Maximum stress versus logarithm of the number of cycles to fatigue failure curve for 2014-T6 Aluminium bar is attached to this solution

Answer:

The maximum allowable load amplitude is 6403.33 N

Step-by-step explanation:

diameter of the bar, d = 12 mm

d = 0.012 m

Cross sectional area of the bar, A = πd²/4

A = π*0.012²/4

A = 0.000113 m²

From the Maximum stress versus logarithm of the number of cycles to fatigue failure for 2014-T6 Aluminium bar attached to this solution;

At 10⁷ cycles, Maximum stress, S = 170 MPa

S = 170 * 10⁶ Pa

Factor of safety, N = 3.0

The tensile stress is given by the formula:

\sigma = S/N\\\sigma = \frac{170 * 10^{6} }{3} \\\sigma = 56.67 * 10^{6} N/m^{2}

The tensile stress is also given by:

\sigma = F/A\\\sigma = F/0.000113

F/0.000113 = 56.67 * 10⁶

F = 56.67 * 10⁶ * 0.000113

F = 6403.33 N

The maximum allowable load amplitude is 6403.33 N

3 0
3 years ago
Read 2 more answers
Will mark Brainliest if someone can help me!!!!! (Idk if Prinicipal is right 2-5)
Talja [164]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1500\\ r=rate\to r\%\to \frac{r}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1,2,3,4,5 \end{cases}

\bf A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 1}\implies A=1500(1.04)^1\implies A\approx 1560\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 2}\implies A=1500(1.04)^2\implies A\approx 1622.4\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 3}\implies A=1500(1.04)^3\implies A\approx 1687.3\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 4}\implies A=1500(1.04)^4\implies A\approx 1754.79\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 5}\implies A=1500(1.04)^5\implies A\approx 1824.98

6 0
3 years ago
Use the number line. Enter the integer value that point D represents.
GenaCL600 [577]

Given:

A number line from -10 to 10 with 20 tick marks.

Point D is 1 tick mark to the left of 5.

To find:

The integer value that represents point D.

Solution:

A number line from -10 to 10 with 20 tick marks. It means, each mark represents the integer values from -10 to 10.

We know that, as we move towards left on a number line the value decreases and as we move towards right the value increases.

Point D is 1 tick mark to the left of 5. It means, point D represents the integer value which is 1 less than 5.

5-1=4

Therefore, point D represents the integer 4.

4 0
2 years ago
this is really simple math, i just forgot how to do this because i haven't done math in so long. help please?
shtirl [24]
X^2 - 2/3x + 1/9   is the answer i belive

3 0
3 years ago
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