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Zolol [24]
3 years ago
11

What is the solution to ther equation 2(x-4)squared=50

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

x=9                   x=-1

Step-by-step explanation:

2(x-4)^2=50

Divide each side by 2

2/2(x-4)^2=50/2

(x-4)^2=25

Take the square root of each side

sqrt((x-4)^2)=sqrt(25)

x-4 = ±5

x-4 =5            x-4 = -5

Add 4 to each side

x-4+4=5+4    x-4+4 = -5+4

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8^6 divided by 8 simplified?? :)
mr_godi [17]

Given: 8^6/8

Answer:32768

Step by step explanation:

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What is the solution
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4 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (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° .

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3 years ago
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