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3 years ago

14

One relaxing attraction at the park is a ferry ride across the river. Visitors are picked up at point A and taken across to poin

t B. The return trip drops them off at point C. Find the length of side AB.

1 answer:

Mumz [18]3 years ago

8 0

<span>If one relaxing attraction at the park is a ferry ride across the river, thus the visitors are picked up at point A and taken across to point B, and the return trip drops them off at point C, then the length of side AB with the endpoints of A is 4,16 and B 20,4 is 7 inches.</span>

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A triangle has sides 42, 4x, and 2x−6. What is the possible range of x? <br><br>? < x < ?

natulia [17]

Answer:

$8\ units< x < 18\ units$

Step-by-step explanation:

we know that

The <u>Triangle Inequality Theorem</u> states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

so

Applying the Theorem

case 1) $42+4x > 2x-6$

$4x-2x > -6-42$

$2x > -48$

$x > -24\ units$

case 2) $42+(2x-6) >4x$

$42-6 >4x-2x$

$36 >2x$

$18 >x$ -----> rewrite

$x$

case 3) $4x+(2x-6) > 42$

$4x+2x > 42+6$

$6x > 48$

$x > 8\ units$

therefore

$8\ units< x < 18\ units$

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3 years ago

I am very confused, if somebody could help me it would mean alot. :)

Yuliya22 [10]

Answer:

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Step-by-step explanation:

3(3)+7=16

9+7=16

16=16

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8 0

3 years ago

If x = y then 3x = 3y gemotry prove

otez555 [7]

Multiplying both sides of an equation by the same number keeps the equation true. In this example, both sides are multiplied by 3, so the equation isn't changed.

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3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

Matthew made a scale model of the solar system for a science project. He created the planet Mercury with a diameter of 6 inches.

Ronch [10]

Answer:

3000

Step-by-step explanation:

I'm learning this right now too!

500×6in=3000

6 0

2 years ago