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Musya8 [376]
3 years ago
14

What is the solution to -4x+5y=14 and 7x+3y=-1 I need the answer now.

Mathematics
2 answers:
Vlad [161]3 years ago
8 0
X= -1 and y=2
Proof:
-4x+5y=14 (equation 1)
7x+3y= -1 (equation 2)

3y= -7x -1
y= -7/3x -1/3 (equation 3)
(3) into (1)
-4x+5(-7/3x-1/3)=14
-4x-35/3x-5/3=14
-47/3x-5/3=14
-47/3x=47/3
x= (-47/3)/ (47/3)
x= -1 
sub x= -1 into (3)
y= -7/3(-1)-1/3
y= 2



Serjik [45]3 years ago
7 0
One way to solve is to try to cancel out some terms
we till try to cancel out the y terms

-4x+5y=14
7x+3y=-1
find lcm of 3 and 5 the lcm=15
-4x+5y=14
multily by 3
-12x+15y=42

7x+3y=-1
multiply both sides by -5
-35x-15y=5
add 2 equations together
-12x-35x+15x-15x=42+5
add like terms
-47x=47
divide both sides by 47
-x=1
multpiply by -1
x=-1

subsitute -1 for x in one quation

7x+3y=-1
x=-1
7(-1)+3y=-1
-7+3y=-1
add 7 to both sides
3y=6
divide both sides by 3
y=2


x=-1
y=2
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Two trains leave the station at the same time, one heading west and the other east. The westbound train travels at 95 miles per
Ludmilka [50]

Answer:

2.2 hr

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8 0
3 years ago
Attached as picture. Please read fully
Troyanec [42]

a. The velocity t = v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. v60 = 7164

<h3>How to solve for the velocity</h3>

mdv/dt = ck - mg

dv/dt = ck/m - mg/m

= ck/m - g

dv = (\frac{ck}{Mo-Kt} -g)dv

Integrate the two sides of the equation to get

v -\frac{ck}{k} e_{n} (Mo- kt)-gt+c

v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. fuel accounts for 55% of the mass

So final mass after fuel is burned out is = 0.45

c=2500

g=9.8

t=60

v = -2500ln0.45 - 9.8 x 60

= 7752 - 588

= 7164

<h3>Complete question</h3>

A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.

dv dt m =ck - mg

(a) Find v(t) keeping in mind that the mass m is a function of t.

v(t) =

m/sec

(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]

v(60) =

m/sec [Round to nearest whole number]

Raed more on velocity here

brainly.com/question/25749514

#SPJ1

8 0
1 year ago
Need help on this please​
lys-0071 [83]

Answer:

I would it would be the one on the left

Step-by-step explanation:

If it did not would then add me and we can talk about it and I will explain to you why it is right.

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