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2 years ago

Twice the larger of two consecutive integers is equal to fifteen less than three times the smaller. Find the integers.

Mathematics

1 answer:

Alexxandr [17]2 years ago

7 0

Answer:

17,18

Step-by-step explanation:

let the smaller integer be "a" and the larger be "b"

Then 2b = 3a -15 ........................equ 1

but they are consecutive integers, b= a+1 ; substitute b into equ 1,thus,

2(a+1) = 3a - 15

2a + 2 = 3a-15

collect like terms,

2+15 =3a-2a

17 = a

b= a+1

b = 17+1

b=18.

The integers are 17,18

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anastassius [24]

Answer:

Step-by-step explanation:

20% of 15 is 3 so Quentin is 18 years old

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The maximum area of a rectangle when given perimeter is simply perimeter/sides.
To get the maximum area, simply divide 160 (perimeter) by 4 (sides in a rectangle) and then square the result:
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Start at 10. Create a pattern that multiplies each number by 10

Yanka [14]

Pattern:

10, 100, 1,000, 10,000, 100,000, 1,000,000, 10,000,000, and so on.

Step-by-step explanation:

First, you must start by using 10 and multiplying it 10 times, 10 x 10 = 100.

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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

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

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