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3 years ago

I need help with all of the odd numbers

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

Hi there!

I will help you by telling you how to solve them. I will show you one of the ways to solve this.

The first way you can solve this is by multiplying and then reducing the fraction. To do this all you need to do is multiply the numerators together and then you multiply the denominators together, after you do this you simply reduce the fraction to it's lowest possible numerator and denominator. For example:

$\dfrac{3}{5} * \dfrac{4}{6} = \dfrac{3*4}{5*6} = \dfrac{12}{30}$

Then to reduce:

$\dfrac{12}{30}$ ÷ $\dfrac{2}{2}$ = $\dfrac{6}{15}$ ÷ $\dfrac{3}{3}$ = $\dfrac{2}{5}$

Hope this helps!

Your friend, ASIAX

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Naily [24]

Answer:

$proj_uw=6.2i-4.2j$

Step-by-step explanation:

The projection of a vector $v$ onto a vector $u$ is defined as the projection of the vector $v$ on the line that contains the vector $u$ . It can be calculated using the following formula:

$proj_uv=\frac{u\cdot v}{||u||^2} u$

Where:

$u\cdot v$

Is the dot product between $u$ and $v$ which is given by:

$u\cdot v= $$\sum_{i=1}^{n} u_iv_i= u_1v_1+u_2v_2+...+u_nv_n$$$

and:

$||u||$

Is the magnitude of vector which can be calculated as follows:

$||u||=\sqrt{u_1^2+u_2^2+...+u_n^2}$

In this sense, the projection of vector w onto vector u is:

$proj_uw=\frac{u\cdot w}{||u||^2} u$

Where the dot product between $u$ and $w$ is:

$u\cdot w =(9*19)+(-6*15)=171-90=81$

And the magnitude of $u$ is:

$||u||=\sqrt{9^2+(-6)^2} = 3 \sqrt{13}$

Thus:

$proj_uw=\frac{u\cdot w}{||u||^2} u=\frac{81}{117} \langle9,-6\rangle=\langle6.23,-4.15\rangle\approx6.2i-4.2j$

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mr_godi [17]

Answer:

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and the center is (-5,-7), so it is + 5 and + 7

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