Find the distance between the two points in simplest radical form.<br> (-5,8) and (-3,1)

Solve the simultaneous equation x+y/4=1/2 and x-3y/3=2

Feliz [49]

Answer:

x+y/4 = 1/2

x-3y/3 = 2

move variables to one side:

multiply the first equation by 4 to get: x+y =2

and the second equation by 3 to get: x-3y =6

then subtract the equations to cancel out x:

x+y = 2

- x-3y = 6

then u get

y--3y = 2-6

4y = -4

y=-1

substitute to solve for x:

x-1 / 4 =1/2

x-1 = 2

x=3

check:

3+-1

2/4= 1/2

correct!!!

3 0

2 years ago

Read 2 more answers

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).

zzz [600]

Answer:

F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

Step-by-step explanation:

We have given $F(S)=\frac{7e^{-7s}}{s^2-49}$

Now $F(S)=e^{-7s}G(s)$

Here $G(S)=\frac{7}{S^2-49}$

Now first find the Laplace inverse of G(S)

Using partial fraction

$\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}$

$7=A(S-7)+B(S+7)$

On comparing the coefficient

$A=\frac{1}{2}$ and $B=\frac{-1}{2}$

On putting the value of A and B

$G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}$

Taking inverse Laplace

$G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}$

Now in G(s) there is onether term $e^{-7s}$

So F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

6 0

2 years ago

A help would be greatly appreciated!

dalvyx [7]

The tree was approximately 40.8ft tall.

You use the equation, a^2+b^2=c^2.
12^2 + 39^2 = x^2, add the results of 12^2 and 39^2 to make 1665 = x^2. You can take the positive root of both sides, 3√185, and it equals 40.80441152, which is rounded to 40.8.

7 0

3 years ago

In the given figure, C is the midpoint of AB and M is the midpoint of AC . Find the length of NB if MC = 4 and AN = 14.

Elina [12.6K]

Answer:

Given MC = 4

AN = 14

To Find, the length of NB

Step-by-step explanation:

AB is a line which has midpoint “C”. Now the line is divided into two equal portion AC and CB.

The AC has midpoint “M” and MC is 4, so AM will also be 4.

N is the midpoint of CB. So, CB = CN + NB

Now we know AC = AM + MC = 4 + 4 =8

Given, AN = 14

AN = AC + CN

14 = 8 + CN

CN = 6

Since N is the midpoint of CB then, CN = NB

Therefore, the NB is 6

8 0

3 years ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation $A=45e^{-0.05t}$ .