Mr. Griffin wants to help his students maintain complete concentration throughout their long calculus exam. To achieve this goal, he should structure the first few questions to: A) Gradually increase in difficulty. B) Cover a wide range of topics. C) Be concise and to the point. D) Include interesting real-life applications.
Answers
Answer:
A. Gradually increase in difficulty
Step-by-step explanation:
Option A is the most effective choice because structuring the first few questions of a long calculus exam to gradually increase in difficulty can help students ease into the exam, build confidence, and maintain their concentration. Starting with easier questions allows students to warm up and gain momentum, which can reduce anxiety and increase their focus. This approach aligns with best practices in assessment and educational psychology, as it promotes a smoother transition into more challenging material, ultimately supporting better concentration and performance throughout the exam.
Final answer:
To maintain student concentration during a long exam, the first few questions should gradually increase in difficulty. This approach builds student confidence and eases them into the problem-solving process, potentially reducing test anxiety and encouraging perseverance through harder problems.
Explanation:
To help his students maintain complete concentration throughout their long calculus exam, Mr. Griffin should structure the first few questions to be gradually increase in difficulty. This approach helps students to gain confidence as they successfully solve the initial questions which is likely to carry them through the rest of the exam and maintain their concentration.
Beginning with easier questions allows the students to 'warm up' and transition their mind into the calculus mode. Then, as the questions become increasingly difficult, students are better prepared to tackle them because they've eased into the problem-solving process instead of being hit with the most challenging problems right off the bat. This approach can reduce test anxiety and encourage perseverance through the more difficult problems towards the end of the test.
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The polynomial 8x2 – 8x + 2 – 5 + x is simplified to 8x2 – gx – h. What are the value of g and h?
b. 69.00 seconds
c. 0.33 seconds
d. 2.30 seconds
Answers
0 = -16t^2 + 85
t = 2.30 seconds
The answer is d.
Answers
#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
Extraneous root in this case is positive four since +4≠-4
In this case it's negative 3, since -3≠3
#29 can be turned into a quadratic equation.
Square both sides to get
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1. Plug both back into the original equation. Whichever does not work is the extraneous root, and the answer is the one that does.
Extraneous root would be negative 3.
Extraneous root would be positive 1.
Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.
Answers
Step-by-step explanation:
2x⁴ = 9x²
2x⁴ - 9x² = 0
x²(2x² - 9) = 0
Either x² = 0 or 2x² - 9 = 0.
When x² = 0, x = 0.
When 2x² - 9 = 0, x² = 9/2, x = ± 3/√2.
Hence the solutions are
x = 0, x = 3/√2 and x = -3/√2.
(b) Determine the domain and the equation of the vertical asymptote.
(c) Make a table of values to find three other points on the graph.
(d) Graph the function. Label the three points you found in Part (c).
Answers
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
-4
4
-8
8
Answers
Your answer is 8
Good morning Brainiac
2-(-6)
2+6
= 8
It was my pleasured to help you :0
Find the volume of the cone.
5
Diameter: 14 m, Slant Height: 25 m
Help Resources
Round to the nearest whole number.
Volume
[?] m3
Answers
The volume of the cone to the nearest whole number is 1283 m³
How to determine the volume
Formula for volume of a cone =
Slant height = 25m
If diameter = 14m , radius = 14/2 = 7m
Pie = 22/7
Substitute values into formula
We have,
Volume =
Volume =
Volume = in the nearest whole number
Thus, the volume of the cone to the nearest whole number is 1283 m³
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Final answer:
The volume of a cone with a diameter of 14 m and slant height of 25 m is 1232 m³, when rounded to the nearest whole number.
Explanation:
To find the volume of the cone, one can use the formula, which is V = 1/3πr²h, where V is the volume, r is the radius, and h is the height. But in the provided case, we have the cone's diameter and slant height instead of the radius and height. Given that the diameter is 14m, the radius would be half of the diameter, so r = 14/2 = 7m. Also, considering the cone as a right-angled triangle, we can use the Pythagorean theorem to find the height. So, h = sqrt((Slant height)² - r²) = sqrt((25)² - (7)²) = 24m. Now, we substitute the values of r and h into the formula for volume of a cone.
V = 1/3 * π * (7)² * 24 = 1232 m³
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