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FinnZ [79.3K]
3 years ago
8

Each side of an equilateral triangle measures 9cm. Find the height, h, of the triangle.

Mathematics
1 answer:
lesantik [10]3 years ago
4 0
Heigh \ for\ equilateral \ triangle \ is \calculated\ from\ formula:\\\\h=\frac{\sqrt3}{2}a\\\\a-side\ of\ triangle\\\\a=9cm\\\\h=\frac{\sqrt3}{2}*9=\frac{9\sqrt3}{2}=4,5\sqrt2cm
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On a coordinate plane, a line is drawn from point a to point b. point a is at (negative 8, negative 13) and point b is at (4, 11
vladimir2022 [97]

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Point a is at (-8, -13) and point b is at (4, 11). Hence the coordinates for point p(x, y) which is one-third the length of the line segment from a to b is:

x=\frac{1}{3}(4-(-8))+(-8)=-4\\ \\y=\frac{1}{3}(11-(-13))+(-13)=-5

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

Find out more on equation at: brainly.com/question/2972832

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2 years ago
I don't care for an explanation PLEASE JUST HELP!!! It would mean a lot !!!! Thanks
Westkost [7]
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On a coordinate plane, a curved line labeled f of x with a minimum value of (1.9, negative 5.7) and a maximum value of (0, 2), c
Bumek [7]

Answer:

4.F(x) > 0 over the intervals (-0.7, 0.76) and (0.76, ∞).

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Caillou received a notice from his bank that his account was overdrawn by $14 he made a deposit that brought his balance to 50.
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Step-by-step explanation:

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Hope this helps!


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~Courtney

6 0
3 years ago
Read 2 more answers
Which real numbers are zeros of the function?
monitta
<h3><u>The roots are -1/2, 0, 2, and 3.</u></h3>

Let's trying factoring this polynomial.

We can factor an x out of each term to start.

x(2x^3 - 9x^2 + 7x + 6)

We now know one of the roots is going to be zero.

Using the rational roots theorem, and the remainder theorem, we can try to find some more roots that way.

Factors of 6: 1, 2, 3, 6.

Factors of 2: 1

Possible rational roots: +/-1/2, +/-1, +/-2, +/-3/2, +/-3, +/-6

Using the remainder theorem, we can plug these values into the polynomial, and if we get a remainder of zero, we know it's a root.

-1/2(2(-1/2)^3 - 9(-1/2)^2 + 7(-1/2) + 6) = 0

Our first root is -1/2.

We can successfully factor a -1/2 out of this polynomial.

After diving the polynomial by -1/2, we're left with: 2x^2 - 10x + 12.

We can now try using the AC method to get our last two roots.

First, we can factor this polynomial to simplify it.

Divide all terms by 2.

2(x^2 - 5x + 6)

Now let's try using the AC method.

The digits -3 and -2 satisfy the criteria.

2(x - 3)(x - 2)

We now have all of our roots: 0, -1/2, 2, and 3.


6 0
3 years ago
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