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3 years ago

Find the lateral area of the square pyramid

Mathematics

1 answer:

fredd [130]3 years ago

5 0

Lateral Area, namely the area of its sides, namely excluding its base.

well, the pyramid is standing on one of its triangular faces, so we'll have to exclude that.

now, let's notice, is a square pyramid, so it has an 8x8 square, and it has triangular faces that have a base of 8 and a height of 22.

$\bf \stackrel{\textit{square's area}}{(8\cdot 8)}~~+~~\stackrel{\textit{3 triangular faces' area}}{3\left[ \cfrac{1}{2}(8)(22) \right]}\implies 64+264\implies 328$

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3 years ago

67. The line contains the point (4,0) and is parallel to the line defined by 3x = 2y.

olganol [36]

Answer:

$y=\frac{3}{2} x-6$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form:

$y=mx+b$ where m is the slope of the line and b is the y-intercept (the value of y when the line crosses the y-axis)

Parallel lines will always have the same slope but different y-intercepts.

1) Determine the slope of the parallel line

Organize 3x = 2y into slope-intercept form. Why? So we can easily identify the slope, m.

$3x = 2y$

Switch the sides

$2y=3x$

Divide both sides by 2 to isolate y

$\frac{2y}{2} = \frac{3}{2} x\\y=\frac{3}{2} x$

Now that this equation is in slope-intercept form, we can easily identify that $\frac{3}{2}$ is in the place of m. Therefore, because parallel lines have the same slope, the parallel line we're solving for now will also have the slope $\frac{3}{2}$ . Plug this into $y=mx+b$ :

$y=\frac{3}{2} x+b$

2) Determine the y-intercept

$y=\frac{3}{2} x+b$

Plug in the given point, (4,0)

$0=\frac{3}{2} (4)+b\\0=6+b$

Subtract both sides by 6

$0-6=6+b-6\\-6=b$

Therefore, -6 is the y-intercept of the line. Plug this into $y=\frac{3}{2} x+b$ as b:

$y=\frac{3}{2} x-6$

I hope this helps!

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Deductive reasoning meanwhile follows that a generally established fact is used to make conclusion about a specific example.

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