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Nana76 [90]
3 years ago
6

Need help with number 24 plz!

Mathematics
1 answer:
morpeh [17]3 years ago
5 0

Answer:its 23{#%]*73

Step-by-step explanation:

Because its that

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the slope of the line will change from negative to positive

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butalik [34]

Answer:

5,720,000 Sq Feet.

Step-by-step explanation:

They are asking the Area of a Trapezoid

Equation =( base + base / 2) x height

Lets first set it up. We know that A-B is 200 feet. & D-C is 260. They are both the bases.

So if you multiply them (200 x 260) You will get 52000.

Now we divide that by 2. (52000 / 2 = 26000)

Now it says the straight line is 220. I think this refers to the height.

Now we multiply 26000 x 220. Which is 5720000.

Its a pretty big number, but that is the answer! :)

Hope this helps!

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Read 2 more answers
What is the estimate of 3630?
Paul [167]
The estimate of 3630 is 4000. because the hundreds place is higher than 5, so you would go a number higher
4 0
3 years ago
a person kicks a ball from the ground into the air with an initial upward velocity of 8 feet per second. what is the maximum hei
KengaRu [80]
Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

\bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}
\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+8t+0

now let's find the y-coordinate of its vertex

\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(\qquad ,~~0-\cfrac{8^2}{4(-16)}  \right)\implies (\quad ,~0+1)\implies (\quad ,~1)

when will it hit the ground?

\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0
\\\\\\
8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}
\\\\\\
2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it  when it came back down}}{t=\cfrac{1}{2}}

3 0
3 years ago
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