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3 years ago

On a parallelogram, the vector from one vertex to another vertex is (9,-2). What is the length of the side?

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

Answer:

The length of the side of the parallelogram is $\sqrt{85}$ .

Step-by-step explanation:

It is given that the vector from one vertex to another vertex is (9,-2), thus we can write it in the vector from that is $v=9\hat{i}+(-2)\hat{j}$ and let the another vertex is adjacent to it.

Now, in order to find the length of the side of the parallelogram, we use the absolute modulus that is $|v|=\sqrt{a^{2}+b^{2}}$ .

Now, since v=(9,-2), thus $|v|=\sqrt{(9)^{2}+(-2)^{2}}$

= $\sqrt{81+4}$

= $\sqrt{85}$

Thus, the length of the side of the parallelogram is $\sqrt{85}$ .

sertanlavr [38]3 years ago

6 0

Answer: Second Option is correct.

Step-by-step explanation:

Since we have given that

In a parallelogram, the vector from one vertex to another vertex is (9,-2)

So, the co-ordinate of first vertex will be (9,0).

And the coordinate of second vertex will be (0,-2).

As we know the formula for "Distance between two coordinates":

So, The length of the side is given by

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\Distance=\sqrt{(9-0)^2+(0-(-2))^2}\\\\Distance=\sqrt{9^2+2^2}\\\\Distance=\sqrt{81+4}\\\\Distance=\sqrt{85}$

Hence, Second Option is correct.

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Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

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Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$