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Masja [62]
4 years ago
5

What is the area of the equilateral triangle? Round to the nearest square centimeter. Do not include units

Mathematics
2 answers:
ziro4ka [17]4 years ago
5 0
The answer should be around 28 square centi. rounding from about 27.71
swat324 years ago
3 0
Area of an equilateral triangle is:

\frac{ \sqrt{3}}{4} 8^{2}=16\sqrt{3}=27.7128≈28

so your answer is 28


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What is the midpoint of the segment shown below?
Bogdan [553]
C. 5/2, 3 this is the answer
3 0
3 years ago
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What is the length of the arc when the degree measure is 60° and the radius is 7?
shepuryov [24]
Inscribed angle =1/2 arc
4 0
3 years ago
If frank has 380 condoms and he uses 180 in two weeks and by the 4th week he has 2 left how many did he use in 2 weeks (btws thi
anastassius [24]

You have said that he used 180 in the first two weeks.

That left him with 200 at the end of the first two weeks.
Then you said that he had 2 left after another 2 weeks.
So during those 2 weeks, he used (200 - 2) = 198 in the 2nd 2 weeks.

What we all expected has not happened at all:  Frank is not slowing down !

But, even so, we have to ask ourselves just what Frank is doing with them all.

Nobody can blame you for wondering.


6 0
4 years ago
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The circumference of a circle is 19π m. What is the area, in square meters? Express your answer in terms of \piπ.
lianna [129]

Answer:

90.25π

Step-by-step explanation:

The circumference of circle is 2πr, or in other words the diameter times π.

The area of a circle is πr^2.

19π is the circumference of this circle.

So, 19 is the diameter, and 9.5 the radius.

The area if (9.5^2)π.

9.5^2=90.25

So, in terms of π, the area of the circle is 90.25π.

7 0
3 years ago
Given f(x) =
sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = \left\{        \begin{array}{ll}            2\cos(\pi x) \text{ for }  x \leq -1 \\ \\          \displaystyle   \frac{2}{\cos(\pi x)}\text{ for } x > -1        \end{array}    \right.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

8 0
3 years ago
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