Answer:
Width of the arch = 105 m
Step-by-step explanation:
Function representing the width of the arch,
f(x) = -0.016(x - 52.5)² + 45
where x = width of the base of the arch or horizontal distance from arch's left end
f(x) = vertical distance of the arch
From the given quadratic function, vertex of the parabola is (52.5, 45).
Coordinates of the vertex represents,
Height of the arch = 45 m
Half of the horizontal distance from the left end = 52.5 m
Therefore, width of the bridge = 2(Half the width of the bridge from left end) = 2×52.5
= 105 m
Therefore, given bridge is 105 m wide.
Can you take a pic of the problem?
Answer:
The second plot.
Step-by-step explanation:
In the picture attached, the scatter plots are shown.
The second plot has minimum residuals and its residuals are randomly distributed. Residual is computed as follows:
residual = measured - predicted
It easy to see that residuals of the first option are greater than the second option.
Line of the second option is better than those from third and fourth options because residuals of the second option are randomly distributed, while in the third option, residuals are mostly negative; and in the fourth option, they are mostly positive.
Answer: 4 (153 - 98)
formula is a^2-b^2=(a+b)(a-b)
factor out the GCF:
4(153−98)
both terms are not perfect squares so
you cant factor it further.
What kind of math is this
