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lukranit [14]
3 years ago
10

If each cube has edges .5 inches long, what is the volume of the prism?

Mathematics
2 answers:
Gre4nikov [31]3 years ago
6 0
.20 is the correct answer
slega [8]3 years ago
5 0
A cube with a side length (S) of .5 has a volume (V) of 0.125
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If 0° ≤ θ ≤ 90° and cosθ = 11 15 , what is the value of sin (90° - θ)?
kolezko [41]
The same value as cosα

cosα=sin(90-α) and vice versa

sinα=cos(90-α)
5 0
3 years ago
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A circle has a diameter of 4 inches. A square has a side length of 4 inches. What is true about the areas of these two figures?
Nuetrik [128]
The area of the square is greater than the circle: the area of the circle is pi(r)^2 while the area of the square is length times width. The area of the circle is about 12.5 while the area of the square is 16.
4 0
3 years ago
More help please!!!!!!!!!!!!!!!!!!!!!!!
Gemiola [76]

Answer:

3 1/3

Step-by-step explanation:

10+z=z+z+z+z

10+z=4z

10=3z

z=3 1/3

8 0
3 years ago
find the equation of the circle where (-9,4),(-2,5),(-8,-3),(-1,-2) are the vertices of an inscribed square.
solniwko [45]
Check the picture below, so, that'd be the square inscribed in the circle.

so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2}
\\\\\\
d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2}
\\\\\\
d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10

that means the radius r = 5.

now, what's the center?  well, the Midpoint of the diagonals, is really the center of the circle, let's check,

\bf \textit{middle point of 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-8-2}{2}~,~\cfrac{-3+5}{2} \right)\implies (-5~,~1)

so, now we know the center coordinates and the radius, let's plug them in,

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}\\
&-5&1&5
\end{array}
\\\\\\\
[x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25

8 0
3 years ago
Lily is a botanist who works for a garden that many tourists visit. The function f(s) = 2s + 30 represents the number of flowers
Vesnalui [34]

Answer:

  A. b(w) = 80w +30

  B. input: weeks; output: flowers that bloomed

  C. 2830

Step-by-step explanation:

<h3>Part A:</h3>

For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...

  b(w) = f(s(w)) = 2(40w) +30

  b(w) = 80w +30 . . . . . . blooms over w weeks

__

<h3>Part B:</h3>

The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.

The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.

Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).

__

<h3>Part C:</h3>

For 35 weeks, the number of flowers that bloomed is ...

  b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks

7 0
2 years ago
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