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iragen [17]
3 years ago
12

Find the volume of the cement block in the figure shown. Please help :)

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0

9514 1404 393

Answer:

  1240 in³

Step-by-step explanation:

The overall dimensions of the block are ...

  10 in by 11 in by 17 in

The volume of that space is ...

  V = LWH = (10 in)(11 in)(17 in) = 1870 in³

The volume of each of the three identical holes is similarly found:

  V = (10 in)(3 in)(7 in) = 210 in³

Then the volume of the block is the overall volume less the volume of the three holes:

  = 1870 in³ - 3(210 in³) = 1240 in³

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The function f(x) = 28(1.45)x models the number of cats and dogs adopted from a local animal shelter, where x represents the num
Phantasy [73]

Answer:

Step-by-step explanation:

If that number is greater than 1, that means that the exponential equation represents growth.  If that number is less than 1 but greater than 0, that means that the exponential equation represents decay.  Since ours is greater than 1, our equation is exponential growth.  It means that the perent of animals adopted grows at a rate of 145% per year, with a starting number of 28 animals in the first year.

6 0
3 years ago
Find the distance of GF if G(9,1) and F (0,2)
galina1969 [7]

Answer:

9.06 units

Step-by-step explanation:

The distance formula for points on a graph is basically the Pythagorean Theorem

√(x2 - x1)² + (y2 - y1)²

Plug in the values and you'll get √82, which equals about 9.05539, or 9.06

7 0
3 years ago
What is the slope of a l8ne that contains the points (34,12) and (32,48)
Vlad [161]
I hope this helps you





slope (34-32)= 12-48




slope.2= -36




slope = -18
7 0
3 years ago
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
What is the common difference for this arithmetic sequence?
OleMash [197]

Answer:

Concept: Series

  1. Start by taking the first and second An and finding the difference
  2. You can then prove your result by taking any ordinary An and subtracting it by An-1
  3. Hence 45-39= 6
  4. To prove our result of 6; 39-33= 6
  5. Hence the common difference is
  6. Rate brainlist
5 0
3 years ago
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