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Paraphin [41]
3 years ago
8

Use the figure to answer the following questions.

Mathematics
1 answer:
Mars2501 [29]3 years ago
7 0
A) AB is where plane P and plane R intersect.
B) A, D, B are collinear (on the same line).
C) plane ADG (three points on the plane)
D) F, D, G, A are all on plane R
E) D lies on both planes.
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Taryn is measuring ribbon to make decorations for a fall party. She needs 12 feet of ribbon. She has 7 and 1/3 feet of ribbon. H
Aliun [14]
Subtract 7.33 from 12.
it gives you 4.67 which in this case can be 4.66 meaning she needs 4 and 2/3 more feet of ribbon, hope this helped
4 0
3 years ago
Find the surface area for both​
Kay [80]
1) 16x16= 256ft
16x3x4= 192ft
add them together
256+192= 448ft2 (2 is the little number at the top next to the unit)

2) 3.14x7(2)= 153.86ft2
and i’m not sure on the other part SORRY
8 0
3 years ago
Determine the sum of the first 20 terms of an arithmetic series with an initial term of 3 and a common difference of 2.
olasank [31]

Answer:

440

Step-by-step explanation:

3 0
4 years ago
A palm tree casts a shadow that is 28 feet long. A 6-foot sign casts a shadow 8 feet long.
s2008m [1.1K]

Answer: The height of the palm tree is 21 feet.

Step-by-step explanation:

We can use a ratio to solve this:

Actual height to shadow for both objects. The fraction equivalents must be equal.

6/8 = x/28 . Cross multiply

6(28) = 8x

168 = 8x   Divide both sides by 8  (8's "cancel" on the right)

168/8 =8x/8

21 = x . This gives us the tree's height as 21 feet.

<em>Another way to solve this is to use the ratios, but simplify the first fraction</em>

<em>6/8 = 3/4</em>

<em>Then multiply the length of the shadow by 3/4</em>

<em>3/4 × 28 = height</em>

<em>28÷4 = 7    7 × 3 = 21</em>

<em>21  feet= the height of the palm tree.</em>

3 0
3 years ago
an electrition charged $70 for 2 hours of work and 115 for 5hours of work write an equtionto find the cost for y and xk
Westkost [7]

\bf \begin{array}{ccll} \stackrel{x}{hours}&\stackrel{y}{cost}\\ \cline{1-2} 2&70\\ 5&115 \end{array}~\hspace{10em}(\stackrel{x_1}{2}~,~\stackrel{y_1}{70})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{115}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{115-70}{5-2}\implies \cfrac{45}{3}\implies 15


\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-70=15(x-2)\implies y-70=15x-30 \\\\\\ y=15x+40\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}


as you already know, a direct proportional variation has a constant of variation "k", y = kx, which in this case that'd be the slope, namely 15.

7 0
3 years ago
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