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yarga [219]
2 years ago
6

Find the surface area of a square pyramid with side length 3 cm and slant height 3 cm

Mathematics
2 answers:
ipn [44]2 years ago
7 0

Answer:

Step-by-step explanation:

base length b = 3 in

slant height L = 3 in

base area = b² = 3² = 9 in²

lateral area = 4·½bL = 4·½·3·3 = 18 in²

surface area = base area + lateral area = 27 in²

Vlada [557]2 years ago
3 0

Answer: 27 cm2

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Let's start with the original function.

y=a sin\frac{2\pi t}{T}

We can immediately fill in the amplitude 'a' and period 'T' , as the question defines these for us, and provides values for 'a' and 'T', 4 and 4\pi respectively.

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Now we only have phase shift and vertical shift to do. Vertical shift is very easy, you can just add it to the end of the right side of the expression. A positive value will shift the graph up, while a negative value will move shift the graph down. We have '-2' as our value for vertical shift, so we can add that on as so:

y=4sin(\frac{2\pit }{4\pi } )-2

Now phase shift the most complicated of the transformations. Basically, it is just movement left or right. A negative phase shift moves the graph right, a positive phase shift moves the graph left (I know, confusing!). Phase shift applies directly to the x variable, or in this case the t variable. To achieve a -4/3 pi phase shift, we need to input +4/3 pi into the function, because of the aforementioned negative positive rule. Here is what the function looks like with the correct phase shift:

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This function has vertical shift -2, phase shift -4/3 \pi, amplitude 4, and period 4\pi.

Desmos.com/calculator is a great tool for learning about how various parts of an equation affect the graph of the function, If you want you can input each step of this problem into desmos and watch the graph change to match the criteria.

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