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3 years ago

How to find the middle of the circle given a point on the circle?

Mathematics

1 answer:

elena-s [515]3 years ago

5 0

Well you need to know a radius or other information

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I need help with number 17

Minchanka [31]

The answer is D complete the square

8 0

3 years ago

Find the base of an isosceles triangle whose area is 60 cm² and the length of one of its equal sides is 13 cm.

Kobotan [32]

Answer:

Base = 24 cm or 10cm

Step-by-step explanation:

REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²

Using the formulas

$A=\frac{bh_b}{2}$

$h_b=\sqrt{a^2-\frac{b^2}{4} }$

There are 2 solutions for $b$

$b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2+\sqrt{13^4-4*60^2} } }$ ≈ $10cm$ $b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2-\sqrt{13^4-4*60^2} } }=24cm$

Less complex:

Area of a triangle = 1/2 * b * h = 60

=> h = 120/b

In right triangle ABD

13² = h² + b² /4 ( by Pythagoras law)

=>169 = 120²/b² + b²/4

=>676 b² = 57600 + b^4

=> b^4 - 676 b² + 57600 = 0

=> b² = 676 +- √(676² - 4*57600) / 2

=> b²= 676 +- √(226576) /2

=> b² = (676 +- 476 )/2

=> b² = 1152/2 , 200 /2

=> b² = 576 , 100

=> b = 24, 10

So, Base = 24 cm or 10cm

4 0

1 year ago

Put me out of my misery! <br> Help me! <br> If possible could you show me how to do this?

prisoha [69]

Answer:

4z+4

Divide each term of 4z+4 by 4 to get 1+z.

1+z

8 0

1 year ago

Read 2 more answers

Write an exponential function to describe the given sequence of numbers.

andriy [413]

Answer:

a(n) = 9·6^(n-1)

Step-by-step explanation:

The general term a(n) of a geometric sequence with first term a(1) and common ratio r is given by ...

a(n) = a(1)·r^(n-1)

Your sequence has first term 9 and common ratio 54/9 = 6, so the function describing it is ...

a(n) = 9·6^(n-1)

5 0

3 years ago

How do I solve the equation in an interval from 0 to 2π ? 9 cos 2t = 6

Ivahew [28]

$9\cos(2t)=6\implies\cos(2t)=\dfrac23$

Using the fact that cos is 2π-periodic, we have

$\cos(2t)=\dfrac23\implies2t=\cos^{-1}\left(\dfrac23\right)+2n\pi$

That is, $\cos(\theta+2n\pi)=\cos\theta$ for any $\theta$ and integer $n$ .

$\implies t=\dfrac12\cos^{-1}\left(\dfrac23\right)+n\pi$

We get 2 solutions in the interval [0, 2π] for $n=0$ and $n=1$ ,

$t=\dfrac12\cos^{-1}\left(\dfrac23\right)\text{ and }t=\dfrac12\cos^{-1}\left(\dfrac23\right)+\pi$

4 0

2 years ago