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3 years ago

Hi everyone I was wondering if someone could please help me out with this problem and explain it to me

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Join the centre O to the chord (let it be MN) & let OH be the perpendicular to the chord

OH bisects MN into 2 equal parts (each one is x/2)
OMH is a right triangle with one side =8, the second leg =x/2 & the hypotenuse = 12 (Radius)
Apply Pythagoras:

12² = 8² +(x/2)² ==>144=64 + x²/4 ==> x²=4(144-64) =320

x²=320==> x=√320 =17.88 ≈17.9

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Tom [10]

Answer: The correct option is (B) f(q) = 2q + 3.

Step-by-step explanation: The given equation representing a function is

$6q=3s-9~~~~~~~~~~~~(i)$

We are to write the equation in function notation with 'q' as independent variable.

From equation (i), we have

$6q=3s-9\\\\\Rightarrow 3s=6q+9\\\\\Rightarrow s=2q+3.$

So, 's' can be represented as a function of 'q'. Hence, we can write

$s=f(q).$

Therefore, we have

$f(q)=2q+3.$

Thus, (B) is the correct option.

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3 years ago

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

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Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

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or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

If AC=5cm, BC=12cm, and m AC= 40 degrees what is the radius of the circumscribed circle

melamori03 [73]

Let's assume they meant C=40 degrees. With an angle like that they're asking for approximation; we'll oblige.

The circumradius is the product of the triangle sides divided by four times the area.

Here we have remaining side given by the Law of Cosines.

$AB^2 = AC^2 + BC^2 - 2\ AC \ BC \cos C$

$AB^2 = AC^2 + BC^2 - 2 AC \ AB \cos C = 5^2 + 12^2 - 2(5)(12) \cos 40^\circ$

$AB = \sqrt{ 169 - 120 \cos 40 ^\circ} \approx 8.77921789$

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The circumradius is $r \approx \dfrac{(5)(12)(8.77921789 )}{ 4 (19.283628) } = 6.829019329$

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3 years ago

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schepotkina [342]

Answer:

im not sure but i think the anwser is its self dont take my anwser its garbadge

Step-by-step explanation:

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3 years ago