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3 years ago

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The expression 1 2 b h 2 1 bhstart fraction, 1, divided by, 2, end fraction, b, h gives the area of a triangle where b bb is t

he base of the triangle and h hh is the height of the triangle. What is the area of a triangle with a base of 7 cm 7 cm7, start text, space, c, m, end text and a height of 4 cm 4 cm4, start text, space, c, m, end text?

1 answer:

adoni [48]3 years ago

4 0

Answer:

14cm²

Step-by-step explanation:

Area of a triangle (A) is 1/2bh where;

b is the base of the triangle

h is the height of the triangle

Given the base = 7cm and height = 4cm,

Substituting the given values into the formula of area of a triangle we have;

A = 1/2×7×4

A = 14cm²

Therefore the area of the triangle is 14cm²

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A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

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$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

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