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ziro4ka [17]
3 years ago
10

Fake answers will be reported

Mathematics
2 answers:
Shkiper50 [21]3 years ago
5 0

Answer:

A. 168.56 in.²

Step-by-step explanation:

28 × 28 = 784 in.²

A = πr²

π · 14²

≈ 615.75 in.²

784 in.² - 615.75 in.²

168.25 in.²

168.25 in.² is close to A. 168.56 in.²

timofeeve [1]3 years ago
3 0

Answer:

A.)

Step-by-step explanation:

I just have a really good feeling. but if you don't think it's A do B because if you separate it it become 13 it's a pattern in math

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First answer = 6/8

Second answer= 5.5/8

Third answer= 6.5/8

Fourth answer= 4/8

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Calculate the area of triangle ABC with altitude CD, given A(−7, −1), B(−1, 5), C(0, 0), and D(−3, 3)
Snezhnost [94]

Answer:

A=25 (Rounded)

Step-by-step explanation:

The first step is to find the length of BA which is the base, then you have the find the length of CD, the hight.

To find the distance, you need to use the distance formula which is

d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} (It does not matter if A is x1 or is B is x1, it just has to remain consistent)

I am going to first find BA

d=\sqrt{\left({-7}-{-1}\right)^{2}+\left({-1}-{5}\right)^{2}}

Next, subtract accordingly

d=\sqrt{\left({-7}+{1}\right)^{2}+\left({-1}-{5}\right)^{2}}

d=\sqrt{\left({-6}\right)^{2}+\left({-6}\right)^{2}}

Once you do that, you want to square -6 and -6 which will give you

d=\sqrt{\left({36}\right)+\left({36}\right)}

Now add 36 with 36 which will leave you with

d=\sqrt{{72}}

Do that same thing for CD. If you did it right, you should be left with

d=\sqrt{{18}}

to find the area of a triangle, you must use the formula

A=\frac{B*H}{2}

The Base is \sqrt{{72}} (BA) and the Hight is \sqrt{{18}} (CD). substitute to get

A=\frac{\sqrt{{72}}*\sqrt{{18}}}{2}

Multiply the Base by the Hight to get

A=\frac{\sqrt{{1,296}}}{2}

Divide 1,296 by 2 to get

A=\sqrt{{648}}

Which is equal to

A=25.455844122715710878430397035775

Rounding it to the nearest whole number you get

A=25

Hope this helps. Feel free to ask any follow-up questions.

Have a good day! :)

6 0
4 years ago
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