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lord [1]
4 years ago
8

Pls send answer pls, I rlly need help thank you so much

Mathematics
1 answer:
shepuryov [24]4 years ago
6 0
Try y=-2x-3 (I think that should be it)
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Find (-10) if () = 8x +5

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Find the angle between the given vectors to the nearest tenth of a degree. u = &lt;-5, -4&gt;, v = &lt;-4, -3&gt; (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

⇒ u.v = 32

Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

⇒ cosx =\frac{32}{5(\sqrt{41})}

⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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