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const2013 [10]
3 years ago
11

The table shows the distance Randy drove on one day of her vacation

Mathematics
1 answer:
horsena [70]3 years ago
4 0
Given the table showing the distance Randy drove on one day of her vacation as follows:

\begin{tabular}
{|c|c|c|c|c|c|}
Time (h)&1&2&3&4&5\\[1ex]
Distance (mi)&55&110&165&220&275
\end{tabular}

The rate at which she travels is given by

rate= \frac{distance}{time}  \\  \\ = \frac{55}{1} =55mi/h

If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by

distance = 55 x 6 = 330 miles.
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The vertex of this parabola is at (2,-1) when the y value is 0 and then x value is 5 what is the coefficient of the squared term
Darina [25.2K]

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}


now, let's expand the squared term to get the standard form of the quadratic.


\bf y=\cfrac{1}{9}(x-2)^2-1\implies y=\cfrac{1}{9}(x^2-4x+4)-1 \\\\\\ y=\cfrac{1}{9}x^2-\cfrac{4}{9}x+\cfrac{4}{9}-1\implies \stackrel{its~coefficient}{y=\stackrel{\downarrow }{\cfrac{1}{9}}x^2-\cfrac{4}{9}x-\cfrac{5}{9}}

4 0
3 years ago
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Please help with geometry homework!!!!
EleoNora [17]

Area of Parallelogram ABCD is equivalent to the area of rectangle AXBY.

Area of rectangle AXBY= Length AX * Length AB= 3 ft * 4√2 ft = 12√2 sq.ft.

Thus Area of Parallelogram is 12√2 sq.ft.

4 0
3 years ago
Someone helppppppppppppp
mihalych1998 [28]

Answer:

Answer is 18

Step-by-step explanation:

18x3=54

54+54=108

18+2=20

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108+40=148

7 0
3 years ago
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I need help plzz!! Just box number 4, TY
nydimaria [60]

12.36% I think sorry if its wrong ;-;


3 0
3 years ago
In ΔLMN, the measure of ∠N=90°, the measure of ∠M=17°, and MN = 20 feet. Find the length of LM to the nearest tenth of a foot.
andrey2020 [161]
20.9 ft
This is a right triangle trigonometry question because N is 90 degrees. MN is adjacent to M and LM is the hypotenuse. Adjacent any hypotenuse use the cosine function. 
cos \theta = \frac{adj}{hyp}
plug in known values
cos(17) = \frac{20}{x}
switch cos(20) and x using the products property
x = \frac{20}{cos(17)}
plug into calculator to get 20.9 ft

3 0
3 years ago
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