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Montano1993 [528]
2 years ago
10

Why in these types of questions(the image) I need to multiply the dominator by itself and by the nominator. and why I can't just

multiply it just by itself.

Mathematics
1 answer:
Lana71 [14]2 years ago
3 0

Answer:

\frac{\sqrt{5x^{2} } }{\sqrt{3x^{3} } } =\sqrt{\frac{5}{3x} }

Step-by-step explanation:

Simply because

A=\frac{5}{3} = \frac{5}{3}*\frac{3}{3}  = \frac{15}{9} =\frac{5}{3}

So its value remains Constant , It's simply as if you're multiplying by 1 which doesn't change anything at all

but when doing this

A=\frac{5}{3} = \frac{5}{3}*\frac{5}{3}  = \frac{25}{9} \neq \frac{5}{3}\\

You changed its value

A now became A^{2}

Hope you understand it

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Answer: sum of all the given three sides

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Find the angle between u = (-2,-5) and v = (5,2)
OleMash [197]

The angle between u = (-2,-5) and v = (5,2) is 134 degrees approximately.

<u>Solution:</u>

Given, two vectors are u = (-2, -5) and v = (5, 2)

We have to find the angle between two vectors.

We know that,

a. b=\|a\| .\|b\| \times \cos \theta

where \theta is angle between vectors a and b

\text { Now vectors are }(-2 \vec{\imath}-5 \vec{\jmath}) \text { and }(5 \vec{\imath}+2 \vec{\jmath})

(-2 \vec{\imath}-5 \vec{\jmath}) \cdot(5 \vec{\imath}+2 \vec{\jmath})=\sqrt{(-2)^{2}+(-5)^{2}} \times \sqrt{5^{2}+2^{2}} \times \cos \theta

\text { since }\|a\|=\sqrt{x^{2}+y^{2}} \text { for } a=x \vec{\imath}+y \vec{\jmath}

\begin{array}{l}{-10-10=\sqrt{29} \times \sqrt{29} \times \cos \theta} \\\\ {-20=29 \times \cos \theta} \\\\ {\cos \theta=\frac{-20}{29}} \\\\ {\theta=\cos ^{-1} \frac{-20}{29}} \\\\ {\theta=133.60}\end{array}

Hence, the angle between given two vectors is 134 degrees approximately.

3 0
3 years ago
There is a line that includes the point (4, 1) and has a slope of 1/4. what is its equation in
Lelechka [254]

Answer:

y=1/4x

Step-by-step explanation:

y=mx+b

m=1/4

so, y=1/4x+b

Now, look at our line's equation so far: . b is what we want, the 1/4 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (4,1).

So, why not plug in for x the number 4 and for y the number 1? This will allow us to solve for b for the particular line that passes through the point you gave!.

(4,1). y=mx+b or 1=1/4 × 4+b, or solving for b: b=1-(1/4)(4). b=0.

y=1/4x+0

3 0
2 years ago
What is the difference between 76.82 and 2.761?<br> 49.210<br> 74.679<br> 74.059<br> 79.581
dsp73
The difference would be 74.059
3 0
2 years ago
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