Hello there.
<span>Which line is a graph of the equation 2x + 5y = 10?
Coordinate plane with the following lines graphed: Line A contains a slope of five-halves and y-intercept of 2; Line B contains a slope of negative five-halves and y-intercept of 2; Line C contains a slope of five-halves and y-intercept of negative 2; and Line D contains a slope of negative five-halves and y-intercept of negative 2.
</span><span>line b</span>
F(x) = -4(x - 2)² + 2
f(x) = -4((x - 2)(x - 2)) + 2
f(x) = -4(x² - 2x - 2x + 4) + 2
f(x) = -4(x² - 4x + 4) + 2
f(x) = -4(x²) + 4(4x) - 4(4) + 2
f(x) = -4x² + 16x - 16 + 2
f(x) = -4x² + 16x - 14
-4x² + 16x - 14 = 0
x = <u>-16 +/- √(16² - 4(-4)(-14))</u>
2(-4)
x = <u>-16 +/- √(256 - 224)</u>
-8
x = <u>-16 +/- √(32)
</u> -8<u>
</u>x = <u>-16 +/- 5.66
</u> -8<u>
</u>x = <u>-16 + 5.66</u> x = <u>-16 - 5.66
</u> -8 -8<u>
</u>x = <u>-10.34</u> x = <u>-21.66</u>
-8 -8
x = 1.2925 x = 2.7075
f(x) = -4x² + 16x - 14
f(1.2925) = -4(1.2925)² + 16(1.2925) - 14
f(1,2925) = -4(1.67055625) + 20.68 - 14
f(1.2925) = -6.682225 + 20.68 - 14
f(1.2925) = 13.997775 - 14
f(1.2925) = -0.002225
(x, f(x)) = (1.2925, -0.002225)
or
f(x) = -4x² + 16x - 14
f(2.7075) = -4(2.7075)² + 16(2.7075) - 14
f(2.7075) = -4(7.33055625) + 43.32 - 14
f(2.7075) = -29.322225 + 43.32 - 14
f(2.7075) = 13.997775 - 14
f(2.7075) = -0.002225
(x, f(x)) = (2.7075, -0.002225)
--------------------------------------------------------------------------------------------
f(x) = 2(x - 2)² + 1
f(x) = 2((x - 2)(x - 2)) + 1
f(x) = 2(x² - 2x - 2x + 4) + 1
f(x) = 2(x² - 4x + 4) + 1
f(x) = 2(x²) - 2(4x) + 2(4) + 1
f(x) = 2x² - 8x + 8 + 1
f(x) = 2x² - 8x + 9
2x² - 8x + 9 = 0
x = <u>-(-8) +/- √((-8)² - 4(2)(9))
</u> <u />2(2)
x = <u>8 +/- √(64 - 72)</u>
4
x = <u>8 +/- √(-8)</u>
4
x = <u>8 +/- √(8 × (-1))</u>
4
x =<u> 8 +/- √(8)√(-1)</u>
4
x = <u>8 +/- 2.83i</u>
4
x = 2 +/- 1.415i
x = 2 + 1.415i x = 2 - 1.415i
f(x) = 2x² - 8x + 9
f(2 + 1.415i) = 2(2 + 1.415i)² - 8(2 + 1.415i) + 9
f(2 + 1.415i) = 2((2 + 1.415i)(2 + 1.415i)) - 16 - 11.32i + 9
f(2 + 1.415i) = 2(4 + 2.83i + 2.83i + 2.00225i²) - 16 - 11.32i + 9
f(2 + 1.415i) = 2(4 + 5.66i + 2.00225) - 16 - 11.32i + 9
f(2 + 1.415i) = 8 + 11.32i + 4.0045 - 16 - 11.32i + 9
f(2 + 1.415i) = 8 + 4.0045 - 16 + 9 + 11.32i - 11.32i
f(2 + 1.415i) = 12.0045 - 16 + 9
f(2 + 1.415i) = -3.9955 + 9
f(2 + 1.415i) = 5.0045
(x, f(x)) = (2 + 1.415i, 5.0045)
or
f(x) = 2x² - 8x + 9
f(2 - 1.415i) = 2(2 - 1.415i)² - 8(2 - 1.415i) + 9
f(2 - 1.415i) = 2((2 - 1.415i)(2 - 1.415i)) - 16 + 11.32i + 9
f(2 - 1.415i) = 2(4 - 2.83i - 2.83i + 2.00225i²) - 16 + 11.32i + 9
f(2 - 1.415i) = 2(4 - 5.66i + 2.00225) - 16 + 11.32i + 9
f(2 - 1.415i) = 8 - 11.32i + 4.0045 - 16 + 11.32i + 9
f(2 - 1.415i) = 8 + 4.0045 - 16 + 9 - 11.32i + 11.32i
f(2 - 1.415i) = 12.0045 - 16 + 9
f(2 - 1.145i) = -3.9955 + 9
f(2 - 1.415i) = 5.0045
(x, f(x)) = (2 - 1.415i, 5.0045)
--------------------------------------------------------------------------------------------
f(x) = -2(x - 4)² + 8
f(x) = -2((x - 4)(x - 4)) + 8
f(x) = -2(x² - 4x - 4x + 16) + 8
f(x) = -2(x² - 8x + 16) + 8
f(x) = -2(x²) + 2(8x) - 2(16) + 8
f(x) = -2x² + 16x - 32 + 8
f(x) = -2x² + 16x - 24
-2x² + 16x - 24 = 0
x = <u>-16 +/- √(16² - 4(-2)(-24))</u>
2(-2)
x = <u>-16 +/- √(256 - 192)</u>
-4
x = <u>-16 +/- √(64)</u>
-4
x = <u>-16 +/- 8</u>
-4
x = <u>-16 + 8</u> x = <u>-16 - 8</u>
-4 -4
x = <u>-8</u> x = <u>-24</u>
-4 -4
x = 2 x = 6
f(x) = -2x² + 16x - 24
f(2) = -2(2)² + 16(2) - 24
f(2) = -2(4) + 32 - 24
f(2) = -8 + 32 - 24
f(2) = 24 - 24
f(2) = 0
(x,f(x)) = (2, 0)
or
f(x) = -2x² + 16x - 24
f(6) = -2(6)² + 16(6) - 24
f(6) = -2(36) + 96 - 24
f(6) = -72 + 96 - 24
f(6) = 24 - 24
f(6) = 0
(x, f(x)) = (6, 0)
<u />
Answer:
A. 2·x² + 16·x + 32 ≥ 254
Step-by-step explanation:
The given dimensional relationship between the dimensions of the photo in the center of the cake and the dimensions of the cake are
The width of the cake = The width of the photo at the center of the cake, x + 4 inches
The length of the cake = 2 × The width of the cake
The area of the cake Wanda is working on ≥ 254 in.²
Where 'x' represents the width of the photo (at the center of the cake), let 'W' represent the width of the cake, let 'L' represent the length of the cake, we get;
W = x + 4
L = 2 × W
Area of the cake, A = W × L ≥ 254
∴ A = (x + 4) × 2 × (x + 4) = 2·x² + 16·x + 32 ≥ 254
The inequality representing the solution is therefore;
2·x² + 16·x + 32 ≥ 254
0.90 Is the answer hello hi ahsiodend
For a parallel line the slope of the lines/equations will be the same but for perpendicular line the slope will be the negative reciprocal
-(1/4)*x+b
b=2-(-1/4)*8
