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2 years ago

For circle H, JN = x, NK = 2, LN = 3, and NM = 6. Solve for x.

Mathematics

1 answer:

Advocard [28]2 years ago

8 0

Answer:

The value of x is 9 ⇒ C

Step-by-step explanation:

In a circle, if two chords intersected at a point inside it there are four segments created, two in each cord, the products of the lengths of the line segments on each chord are equal.

<em>Let us use this rule to solve the question</em>

In circle H

∵ Jk and LM are chords intersected at point N inside the circle

∴ The created segments are JN, Nk, and LN, NM

→ By using the rule above

∴ JN × NK = LN × NM

∵ JN = x and NK = 2

∵ LN = 3 and NM = 6

→ Substitute these values in the equation above

∴ x × 2 = 3 × 6

∴ 2x = 18

→ Divide both sides by 2 to find x

∴ $\frac{2x}{2}=\frac{18}{2}$

∴ x = 9

∴ The value of x is 9

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2 years ago

A 100-gallon barrel, initially half-full of oil, develops a leak at the bottom. Let A(t) be the amount of oil in the barrel at t

gayaneshka [121]

Answer:

the mathematical model is : $-\frac{1}{A}= kt - \frac{1}{50}$

Step-by-step explanation:

Given that:

Let A(t) to be the amount of oil in the barrel at time t.

However; Suppose that the amount of oil is decreasing at a rate proportional to the product of the time elapsed and the amount of oil present in the barrel.

Then,

$\frac{dA}{dt} \ \alpha \ A^2$

$\frac{dA}{dt}= KA^2$

Initially the 100 -gallon barrel is half-full of oil

So, A(0) = 100/2 = 50

$\frac{dA}{dt}= KA^2 \ \ \ \ \ \ :A(0)=50$

The variable is now being separated as:

$\frac{dA}{A^2}=kdl$

Taking integral of both sides; we have:

$\int\limits\frac{dA}{A^2}=\int\limits \ kdt$

$-\frac{1}{A}= kt +C$

However; since A(0) = 50; Then

t = 0 ; A =50 in the above equation

$-\frac{1}{50}= 0 +C$

$C = - \frac{1}{50}$

Thus; the mathematical model is : $-\frac{1}{A}= kt - \frac{1}{50}$

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2 years ago

The marginal price per pound (in dollars) at which a coffee store is willing to supply x pounds of Jamaican Blue Mountain coffee

il63 [147K]

Answer:

Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

Step-by-step explanation:

We are given the following information in the question:

The marginal price per pound (in dollars) is given by:

$p'(x) = \displaystyle\frac{208}{(x+7)^2}$

where x is the supply in pounds.

$P(x) = \displaystyle\int p'(x)~dx =\displaystyle\int\displaystyle\frac{208}{(x+7)^2}~dx\\\\P(x) = \frac{-208}{(x+7)} + c\\\\\text{where c is the constant of integration.}$

The coffee shop is willing to supply 9 pounds per week at a price of $7 per pound.

Thus, we are given that

P(9) = 7

Putting the values, we get,

$P(x) = \displaystyle\frac{-208}{(x+7)} + c\\\\P(9) = 7\\\\\displaystyle\frac{-208}{(9+7)} + c = 7\\\\c = 7 + \frac{208}{16} = 20$

$P(x) = \displaystyle\frac{-208}{(x+7)} + 20$

Now, we have to find how many pounds it would be willing to supply at a price of $4 per pound.

P(x) = 4

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Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

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2 years ago

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Step-by-step explanation:

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The equation to represent the costs would be the same is given by :_

$C(x)=c(x)\\\Rightarrow50x+250=75x+100$

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