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NeX [460]
3 years ago
15

Divide.

Mathematics
1 answer:
joja [24]3 years ago
5 0
I know that the answer is 318
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for v= 4i - 5j, find unit vector u in the direction of v, and write your answer as a linear combination of the standard unit vec
umka21 [38]

Answer:

a. u=\frac{4\sqrt{41}i }{41}-\frac{5\sqrt{41}j}{41}

Step-by-step explanation:

The given vector is v= 4i - 5j

The magnitude of this vector is;

|v|=\sqrt{(-4)^2+(-5)^2}

|v|=\sqrt{16+25}

|v|=\sqrt{41}

The unit vector u in the direction of v is;

u=\frac{v}{|v|}

u=\frac{4i - 5j}{\sqrt{41}}

u=\frac{4i }{\sqrt{41}}-\frac{5j}{\sqrt{41}}

We rationalize to get

u=\frac{4\sqrt{41}i }{41}-\frac{5\sqrt{41}j}{41}

4 0
3 years ago
Read 2 more answers
Plz help i do not understand the question.
gavmur [86]
The question is asking you to come up with a real-life example of something that can be represented as 15+ c= 17.50. If I were you I would do sales tax. You bought an item with a 15 dollar price tag, but when you go to check out 2.50 are added in tax to equal 17.50. to solve subtract 15 from 17.50: c= 2.50.  Your variable (c) represents the sales tax.  Hope that helps! :)
6 0
4 years ago
1/4z-2/7=5/7 what is z
Alenkasestr [34]

Answer:

z=4

Step-by-step explanation:

1/4z-2/7=5/7

1/4z= 5/7+2/7

1/4z=1

z=4

3 0
3 years ago
When given a radical to simplify, if no number appears in the index location then it is a one.
slamgirl [31]

Answer:

that is false! im pretty sure!

let me know if u need any more help cutie!

Step-by-step explanation:

6 0
3 years ago
What are the potential solutions of In(x^2-25)=0?
Maslowich

Answer:

x =\pm \sqrt{26}

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

  • ln ( x² - 25 ) = 0

And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,

<u>Equation</u><u> </u><u>:</u><u>-</u><u> </u>

\implies log_e {(x^2-25)}= 0

<u>In </u><u>general</u><u> </u><u>:</u><u>-</u><u> </u>

  • If we have a logarithmic equation as ,

\implies log_a b = c

Then this can be written as ,

\implies a^c = b

In a similar way we can write the given equation as ,

\implies e^0 = x^2 - 25

  • Now also we know that a^0 = 1 Therefore , the equation becomes ,

\implies 1 = x^2 - 25 \\\\\implies x^2 = 25 + 1 \\\\\implies x^2 = 26 \\\\\implies x =\sqrt{26} \\\\\implies x = \pm \sqrt{ 26}

<u>Hence</u><u> the</u><u> </u><u>Solution</u><u> </u><u>of </u><u>the</u><u> given</u><u> equation</u><u> is</u><u> </u><u>±</u><u>√</u><u>2</u><u>6</u><u>.</u>

6 0
3 years ago
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