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tigry1 [53]
3 years ago
9

What is the answer to solve -2=4r+s for s

Mathematics
1 answer:
sleet_krkn [62]3 years ago
7 0
The answer is s= -(4r+2)
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Given the conditional statement below, what is the hypothesis? If you do your homework, then you will pass.​
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The hypothesis is that by doing your homework for the class you will pass

Step-by-step explanation:

If you do your homework, you will pass the class

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Three consecutive intergers have a sum of 54. Find the integers
Aleksandr [31]
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They add up to 54.

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Determine whether the following graphs represent a function by using the vertical line test
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A function passes the vertical line test when it doesn't hit the same spot twice. If I am correct the answers are below.

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List the 4 factors of 8
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3 0
3 years ago
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(1 point) If p(x) and (x) are arbitrary polynomials of degree at most 2, then the mapping =p(-1)q(-1) + p(070) +p(3)q(3) defines
Ainat [17]

If p(x) and q(x) are arbitrary polynomials of degree at most 2 then

||p||||q|| = 26(\sqrt{640}) and angle between p(x) and q(x) is 0.233.

Given that

<p,q> = p(-1)q(-1) + p(0) q(0) + p(3)q(3)

and p(x) = 2x²+ 6 , q(x)= 4x²-4x

then the values of p and q at x = -1,0,3 are given as;

x = -1,

p(-1) = 2(-1)² + 6 = 8   ,    q(-1) = 4(-1)² - 4(-1) = 8

x = 0,

p(0) = 2(0)² + 6 = 6   ,    q(0) = 4(0)² - 4(0) = 0

x = 3,

p(3) = 2(3)² + 6 = 24  ,  q(3) = 4(3)²- 4(3) = 24.

<p,q> = p(-1)q(-1) + p(0)q(0) + p(3)q(3)

         = (8)(8) + (6)(0) + 24(24)

         = 64 + 0 + 576

<p,q> = 640

Now we have to find ||p|| ||q||, for this we'll find ||p|| and ||q||

||p|| = \sqrt{ < p,p > }

     = \sqrt{8(8) + 6(6) + 24(24)}

     = \sqrt{676}

||p|| = 26

and

||q|| = \sqrt{ < q,q > }

      =\sqrt{8(8) + 0(0) + 24(24)}

||q||  =\sqrt{640}

∴||p||||q|| = 26(\sqrt{640\\)

Now we have to find angle between p(x) and q(x),

∴ α = cos⁻¹\frac{ < p,q > }{||p||||q||}

      = cos ⁻¹ \frac{640}{26(\sqrt{640}) }

      = cos ⁻¹ \frac{4\sqrt{10} }{13}

  α  = 13.34°

In radian

α = 0.233.

To know more about Inner product here

brainly.com/question/14185022

#SPJ4

5 0
1 year ago
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