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Anettt [7]
3 years ago
11

You want to improve your grades so that you make all A's next 6 weeks. List examples of quantitative and qualitative data you sh

ould collect. Suggest at least TWO ways to apply each CT strategy to the problem GIVE EXAMPLES and DETAILS!!!!!!!! You DO NOT simply define Plz help I will give brainliest
Engineering
1 answer:
Naddika [18.5K]3 years ago
6 0

Explanation:

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3fभथठभदाफमदखज्ञफादफज्ञादफज्ञिलफ इऋबिअऋब

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What is a chipping hammer used for? <br><br> State three things.
aleksley [76]

Answer:

i hope this helps.

Explanation:

they are used for breaking concrete, can be positioned to break vertical and overhead surfaces, allows precisely chip away only specific areas.

7 0
3 years ago
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"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.  

6 0
3 years ago
If an imbalance occurs, the _
pochemuha

A. AFGI is the answer for this question.

7 0
3 years ago
What do you guys like in engineering
Drupady [299]

Answer:

building lol and actually workin

Explanation:

3 0
3 years ago
Read 2 more answers
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
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