Answer:
-4
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
4x + 5y = -12
-2x + 3y = -16
<u>Step 2: Rewrite Systems</u>
-2x + 3y = -16
- Multiply everything by 2: -4x + 6y = -32
<u>Step 3: Redefine Systems</u>
4x + 5y = -12
-4x + 6y = -32
<u>Step 4: Solve for </u><em><u>y</u></em>
<em>Elimination</em>
- Combine 2 equations: 11y = -44
- Divide 17 on both sides: y = -4
Answer:
8y^3 + 6y^2 - 29y + 15
Step-by-step explanation:
8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15
then
8y^3 + 6y^2 - 29y + 15 is the product
Best regards
I am thinking of a rectangle that has the two sides parallel to one another. Set the two functions equal to one another, so
9x-14=7x+4
After a bunch of algebra and math magic, 2x=18 => x=9
So if you just insert 9 into both equations, both will end up with a value of 67, so it ends up looking like a right triangle. Don't do that. Instead, to find the rectangle widths, use 9x-4 (+10 added to intercept) instead, while keeping 7x+4, so that the intercepts match.
LN) 9x-4 = 9(9)-4 = 81-4 = 77
MP) 7(9)+4 = 63+4 = 67
*If you are also looking for the diagonals, use Pythagorean Theorem 77^2+67^2=(hypotenuse)^2*
Please, write "x^3" for "the cube of x," not "x3." "^" denotes exponentiation.
Then you have g(x) = x^3 - 5 and (I assume) h(x) = 2x - 2.
1) evaluate g(x) at x = -2: g(-2) = (-2)^3 - 5 = -8 - 5 = -13
2) let the input to h(x) be -13: h(-13) = 2(-13) - 2 = -28 (answer)
