Answer: It is TRUE. A point in a linear equation can be the solution to said equation.
It looks like your equations are
7M - 2t = -30
5t - 12M = 115
<u>Solving by substitution</u>
Solve either equation for one variable. For example,
7M - 2t = -30 ⇒ t = (7M + 30)/2
Substitute this into the other equation and solve for M.
5 × (7M + 30)/2 - 12M = 115
5 (7M + 30) - 24M = 230
35M + 150 - 24M = 230
11M = 80
M = 80/11
Now solve for t.
t = (7 × (80/11) + 30)/2
t = (560/11 + 30)/2
t = (890/11)/2
t = 445/11
<u>Solving by elimination</u>
Multiply both equations by an appropriate factor to make the coefficients of one of the variables sum to zero. For example,
7M - 2t = -30 ⇒ -10t + 35M = -150 … (multiply by 5)
5t - 12M = 115 ⇒ 10t - 24M = 230 … (multiply by 2)
Now combining the equations eliminates the t terms, and
(-10t + 35M) + (10t - 24M) = -150 + 230
11M = 80
M = 80/11
It follows that
7 × (80/11) - 2t = -30
560/11 - 2t = -30
2t = 890/11
t = 445/11
It makes and obtuse right triangle.
Answer:
- table: 14, 16, 18
- equation: P = 2n +12
Step-by-step explanation:
Perimeter values will be ...
rectangles . . . perimeter
1 . . . 14
2 . . . 16
3 . . . 18
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The perimeter of a figure is twice the sum of the length and width. Here, the length is a constant 6. The width is n, the number of rectangles. So, the perimeter is ...
P = 2(6 +n) = 12 +2n
Your equation is ...
P = 2n +12 . . . . . . . . perimeter P of figure with n rectangles.
_____
<em>Additional comment</em>
You may be expected to write the equation using y and x for the perimeter and the number of rectangles. That would be ...
y = 2x +12 . . . . . . . . . perimeter y of figure with x rectangles
-4 is Less than -3 is the answer
