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3 years ago

I need help on all of this problems please help

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

If we are working in a coordinate plane where the endpoints has the coordinates (x1,y1) and (x2,y2) then the midpoint coordinates is found by using the following formula:

midpoint=(x1+x22,y1+y22)

Step-by-step explanation:

You might be interested in

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

Giving loads of points to anyone who can answer these questions or even just most of them correctly!!!

cestrela7 [59]

Answer:

below :)

Step-by-step explanation:

1) 12

2}3/2

3)224.64

4)2 quarters

5)w=50

6)m=6y+7

7)42/6=7 x=6

8 0

2 years ago

How many solutions are there to the equation below 9x + 27 = 9( x + 2 ) + 9

Nataly [62]

Let's solve the equation:

9x+27 = 9(x+2)+9 ← Distribute 9 to the x and 2

9x+27 = 9x+18+9 ← Combine like terms

9x+27 = 9x + 27 ← Subtract 27 from both sides

9x = 9x

Infinitely many solutions would be correct because no matter what x is, it will always equal each other the both sides of the equation because it is 9 times x on both sides.

7 0

3 years ago

Read 2 more answers

Many insects migrate (travel) between their summer and winter homes. The desert locust migrates about 800 miles farther than the

Harman [31]

Answer:

Distance travelled by Monarch butterfly = 2000 miles

Distance travelled by desert locust = 2800

Distance travelled by pink-spotted hawk moth= 7800 miles

Step-by-step explanation:

Let the distance travelled by Monarch butterfly be "X" miles

Distance travelled by desert locust is equal to X + 800 miles

Distance travelled by pink-spotted hawk moth is equal to 4X -200 miles

The total distance travelled by the three is equal to 12600 miles

$X + X + 800 + 4X -200 = 12600\\6X = 12600 -600\\6X = 12000\\X = 2000$

Distance travelled by Monarch butterfly = 2000 miles

Distance travelled by desert locust = 2000 + 800 = 2800

Distance travelled by pink-spotted hawk moth = 4 * 2000 -200 = 7800 miles

6 0

3 years ago

Solving these quadratics for x using complete the square 3x^2-4x-1=0. Show your work.

UkoKoshka [18]

Answer: x = (sqrt(7) + 2)/3 and
x = ( – sqrt(7) + 2)/3

Explanation:

3x^2 - 4x - 1 = 0

Divide both sides by 3:

3x^2/3 - 4/3x - 1/3 = 0/3
x^2 - 4/3x - 1/3 = 0
x^2 - 4/3x = 1/3
x^2 - 4/3x + (2/3)^2 = 1/3 + (2/3)^2
(x - 2/3)^2 = 1/3 + 4/9
(x - 2/3)^2 = 7/9

Sqrt both sides:

x - 2/3 = sqrt (7/9)
x - 2/3 = |sqrt(7)/3|

Set x -2/3 = sqrt(7)/3

=> x = (sqrt(7) + 2)/3

Set x - 2/3 = - sqrt(7)/3

=> x = ( - sqrt(7) + 2)/3

6 0

3 years ago

I need help on all of this problems please help ​

I need help on all of this problems please help