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3 years ago

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.

Mathematics

1 answer:

Lemur [1.5K]3 years ago

7 0

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)

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Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

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4 years ago

Find the <br> (D)<br> (R)<br> &<br> (A)

omeli [17]

<h3>Answer: Choice C</h3>

domain = (-infinity, infinity)
range = (-infinity, 0)
horizontal asymptote is y = 0

========================================================

Explanation:

Since no division by zero errors are possible, and other domain restricting events are possible, we can plug in any x value we want. This means the domain is the set of all real numbers. Representing this in interval notation would be (-infinity, infinity).

The range is the set of negative real numbers, which when written in interval notation would be (-infinity, 0). This is because y = 5^x has a range of positive real numbers, and it flips when we negate the 5^x term. The graph of y = -5^x extends forever downward, and the upper limit is y = 0.

It never reaches y = 0 itself, so this is the horizontal asymptote. Think of it like an electric fence you can get closer to but can't touch.

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3 years ago

A trough is filled with water. The trough holds 315 gallons. Each cubic foot of water contains about 7.5 gallons. The trough is

elena55 [62]

Answer:

The height of the trough is about 1.5 ft

Step-by-step explanation:

If each cubic foot of water contains about 7.5 gallons.

Then; 315 gallons is about $\frac{315}{7.5}=42ft^3$

Let h be the height of the trough, then

$7\times 4\times h=42$

This implies that;

$28h=42$

Divide both sides by 28 to get:

$h=\frac{42}{28}$

$\therefore h=1.5$

The height of the trough is about 1.5 ft

4 0

3 years ago

I need to know how to simplify this equation

ArbitrLikvidat [17]

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4 years ago

Can abyone please give the answer to these questions?

Nady [450]

I don't know how to solve 11 but I guess I can solve 12 so I think those are the answers:

12.

(4, 53)

(5, 85)

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3 years ago

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.​

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.