18+11v=w-13t
18+11-w=-13t
13t=w-11v-18
t=1/13w - 11/13v - 1 5/13
Hope this helps :)
Answer:
(You only need to give one solution)
Step-by-step explanation:
We have the following equation
First, we need to foil out the parenthesis
Now we can combine the like terms
Now, we need to factor this equation.
To factor this, we need to find a set of numbers that add together to get -3 and multiply to give us -4.
The pair of numbers that would do this would be 1 and -4.
This means that our factored form would be
As the first binomial is a difference of squares, it can be factored futher into
Now, we can get our solutions.
The first binomial will produce two complex (Not real) solutions.
So our solutions to this equation are
The solution to the first expression - 7+3(9-4)^2÷5 is given as 22.
To get the answer correctly, one must follow rudimentary rules of operations which are coined into the acronym BODMAS.
<h3>What is BODMAS?</h3>
This is the order in which mathematical operations must be executed.
B = Bracket
O = Orders (that is Powers, Indices or roots)
D= Division
M = Multiplication
A = Addition
S = Subtraction
Now lets see how we got 22 from the first set of operations:
<h3>Operation 1 (Example)</h3>
7+3(9-4)^2÷5 =
7+3 (5)^2÷5=
7+3 * 25÷5 =
7+3*5=
7+15=
22
Following the BODMAS rule and the example in Operation 1 above, we can state the remaining answers as follows:
<h3>
Operation 2</h3>
12/3-4+7^2 = 49
<h3 /><h3>
Operation 3</h3>
(7-3)×3^3÷9 = 12
<h3>Operation 4</h3>
5(7-3)^2÷(6-4)^3-9 = 1
<h3>Operation 5</h3>
3×(7-5)^3÷(8÷4)^2-5 = 1
<h3>Operation 6</h3>
9+(3×10)/5×2-12 = 9
See the link below for more about Mathematical Operations:
brainly.com/question/14133018
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
