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goblinko [34]
3 years ago
6

What is T = 3U/E solve for U

Mathematics
1 answer:
rjkz [21]3 years ago
6 0
<span>To solve for U in the equation T = 3U/E, you need to isolate U alone on one side of the equation. To do this, first multiply both sides of the equation by E, to get TE= 3U. Then divide both sides of the equation by 3 to get TE/3 = U. This provides the correct answer: U = TE/3.</span>
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Which equation is NOT a linear function?
Stells [14]

Answer:

y=x^2-2

Step-by-step explanation:

It should be this one because if you have x squared it will be a parabola. A line should have a consistent slope. so Somethingx + y intercept =y

6 0
3 years ago
The distance between A (-4, 2) and B(4, 2) is.<br> answers;<br> a - 6<br> b - 8<br> c - 4
Fofino [41]
B-8. This is because -4 to 0 is 4 and then 0-4 is 4. 4+4=8
6 0
2 years ago
4x+y+2z=4<br> 5x+2y+z=4<br> x+3y=3
vekshin1

Objective: Solve systems of equations with three variables using addition/elimination.

Solving systems of equations with 3 variables is very similar to how we solve systems with two variables. When we had two variables we reduced the system down

to one with only one variable (by substitution or addition). With three variables

we will reduce the system down to one with two variables (usually by addition),

which we can then solve by either addition or substitution.

To reduce from three variables down to two it is very important to keep the work

organized. We will use addition with two equations to eliminate one variable.

This new equation we will call (A). Then we will use a different pair of equations

and use addition to eliminate the same variable. This second new equation we

will call (B). Once we have done this we will have two equations (A) and (B)

with the same two variables that we can solve using either method. This is shown

in the following examples.

Example 1.

3x +2y − z = − 1

− 2x − 2y +3z = 5 We will eliminate y using two different pairs of equations

5x +2y − z = 3

1

3x +2y − z = − 1 Using the first two equations,

− 2x − 2y +3z = 5 Add the first two equations

(A) x +2z = 4 This is equation (A), our first equation

− 2x − 2y +3z = 5 Using the second two equations

5x +2y − z = 3 Add the second two equations

(B) 3x +2z = 8 This is equation (B), our second equation

(A) x +2z = 4 Using (A) and (B) we will solve this system.

(B) 3x +2z = 8 We will solve by addition

− 1(x +2z) =(4)( − 1) Multiply (A) by − 1

− x − 2z = − 4

− x − 2z = − 4 Add to the second equation, unchanged

3x +2z = 8

2x = 4 Solve, divide by 2

2 2

x = 2 We now have x! Plug this into either(A) or(B)

(2) +2z = 4 We plug it into (A),solve this equation,subtract 2

− 2 − 2

2z = 2 Divide by 2

2 2

z = 1 We now have z! Plug this and x into any original equation

3(2) +2y − (1)= − 1 We use the first, multiply 3(2) =6 and combine with − 1

2y + 5= − 1 Solve,subtract 5

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(2, − 3, 1) Our Solution

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5 0
3 years ago
Cole owns a corner store that resells pastries from a local bakery. The table below shows the costs charged for a week of orderi
chubhunter [2.5K]

Let:

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5 0
1 year ago
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elena55 [62]

Answer:

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Quadrant I      -          0° to 90°

Quadrant II     -       90° to 180°

Quadrant III    -      180° to 270°

Quadrant IV    -      270° to 360°

Therefore 20 degree is in First Quadrant, i.e Quadrant I.

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Quadrant IV  -          0° to -90°

Quadrant III   -       -90° to -180°

Quadrant II    -      -180° to -270°

Quadrant I     -      -270° to -360°

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