X -y = 3 solve for y

Mathematics

2 answers:

Wittaler [7]3 years ago

4 0

Answer:

y=x−3

Step-by-step explanation:

x−y=3

Step 1: Add -x to both sides.

x−y+−x=3+−x

−y=−x+3

Step 2: Divide both sides by -1.

−y/−1=−x+3/1

y=x−3

nirvana33 [79]3 years ago

3 0

Answer:

y= -3+x

Step-by-step explanation:

hope this helps :)

You might be interested in

3x2 – x – 4 factorise

never [62]

This factors out to: (3x−4)(x+1) :)

4 0

3 years ago

REALLY NEED HELP ON THESE!!

nata0808 [166]

The answer would be 56n^2 + 19n - 15

Multiply 8n with 7n to get 56n^2, then multiple 8n with 5 and you get 40n, then multiple -3 with 7n to get -21n, then -3 times 5 which gives you -15. Simplify and that is the answer you would get.

6 0

4 years ago

The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func

alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

$h(t)=-16t^2 + 96t$

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$