Find the x-component of this vector:

  -13.9

This value is approximate.

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Explanation:

This angle is in quadrant Q2, which is where x is negative and y is positive.

The angle shown of 65 degrees is added to the 90 degree angle to the right resulting in 65+90 = 155 degrees overall. See diagram below.

The x component of the vector is found by the formula

x = r*cos(theta)

we will plug in r = 15.3 as the magnitude of the vector and theta = 155 as the angle

x = 15.3*cos(155)

x = -13.8665 approximately

x = -13.9 rounding to one decimal place (as 15.3 and 65.0 are both rounded to one decimal place)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2