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3 years ago

An equilateral triangle has an altitude of 12.84 m. What is the length of the three equal sides?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

4 0

Answer:

14.83m

Step-by-step explanation:

In an equilateral triangle, of side l and altitude h we have:

h=l/2×sqrt(3).

l=2h/sqrt(3)=2×12.84/1.73=14.83m

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If tan = 12/5 and cos = -5/13, then what is sin?

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Answer:

sin -12/13

Step-by-step explanation:

You can use the identity ...

tan = sin/cos

to solve for sine:

sin = cos·tan

sin = (-5/13)(12/5) =

sin -12/13

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3 years ago

Solve S = 2tr(h + r) for h.

Ivanshal [37]

Answer:

Step-by-step explanation:

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3 years ago

◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

So the answer would be A.

3 0

3 years ago

Plz answer my question it is urgent..!!

oee [108]

Answer:

In the step-by-step explanation!

Step-by-step explanation:

Not sure if it is too late but here:

1.)

$\frac{1}{\alpha}+\frac{1}{\beta} \\\frac{\beta}{\alpha \beta } +\frac{\alpha}{\alpha \beta } \\\frac{\alpha +\beta }{\alpha \beta }$