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4 years ago

What is the teacher's purpose behind giving short answer tests? (site 1)

SAT

2 answers:

Step2247 [10]4 years ago

7 0

just to see what you know

snow_tiger [21]4 years ago

7 0

Answer:

Short-answer tests are composed of items that are similar to objective items, in that a clearly-defined answer is required. They differ from the latter in that the answer has to be supplied by the person being tested rather than simply chosen from a number of options provided. A teacher's primary purpose in giving a short-answer test is to test whether you have a foundation of knowing the material, usually factual. The main reason teachers test students is to measure student learning and growth (increased understanding and/or skill). Tests also provide feedback on the effectiveness of the teacher's methods and lesson materials. If most students perform well - it means that the teacher's methods are effective.

Explanation:

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3 0

3 years ago

Jerry writes down all the odd numbers $1, 3, 5, 7, \dots,$ up to $999$. what is the sum of all the numbers he writes down?

Pavel [41]

The sum of n odd numbers n², where n is a natural number and represents the number of terms.

<h3>What is a natural number?</h3>

The numbers that we use when we are counting or ordering, which occurs commonly and obviously in nature.

Here, given all the odd numbers from 1,3,5,7..........999.

The sum of n odd numbers, n².

999² = 998001

Thus, the sum of all the odd numbers from 1 to 999 is 998001.

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7 0

2 years ago

The point (−2,4) lies on the curve in the xy-plane given by the equation f(x)g(y)=17−x−y, where f is a differentiable function o

Nikitich [7]

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

<h2>Procedure - Differentiability</h2><h3 /><h3>Chain rule and derivatives</h3><h3 />

We derive an expression for $\frac{dy}{dx}$ by means of chain rule and differentiation rule for a product of functions:

$\frac{d}{dx}[f(x)\cdot g(y)] = [f'(x)\cdot \frac{dx}{dx}]\cdot g(y) + f(x) \cdot [g'(y)\cdot \frac{dy}{dx} ]$

$\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx}$ (1)

If we know that $f(x) \cdot g(y) = 17-x-y$ , $f(-2) = 3$ ,<em> </em> $f'(-2) = 4$ ,<em> </em> $g(4) = 5$ and $g'(4) = 2$ , then we have the following expression:

$-1-\frac{dy}{dx} = (4)\cdot (5) + (3)\cdot (2) \cdot \frac{dy}{dx}$

$-1-\frac{dy}{dx} = 20 + 6\cdot \frac{dy}{dx}$

$7\cdot \frac{dy}{dx} = -21$

$\frac{dy}{dx} = -3$

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

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<h3>Remark</h3>

The statement is incomplete and full of mistakes. Complete and corrected form is presented below:

<em>The point (-2, 4) lies on the curve in the xy-plane given by the equation </em> $f(x)\cdot g(y) = 17 - x\cdot y$ <em>, where </em> $f$ <em> is a differentiable function of </em> $x$ <em> and </em> $g$ <em> is a differentiable function of </em> $y$ <em>. Selected values of </em> $f$ <em>, </em> $f'$ <em>, </em> $g$ <em> and </em> $g'$ <em> are given below: </em> $f(-2) = 3$ <em>, </em> $f'(-2) = 4$ <em>, </em> $g(4) = 5$ <em>, </em> $g'(4) = 2$ <em>. </em>