Answer:
the third one is the correct answer
we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
<h3>How many more inches of snow were on the ground at 5:00 p.m. than at 8:00 a.m.?</h3>
We know that at 8:00 a.m. there were t inches of snow in the ground.
At 5:00 p.m. there were 3t inches of snow in the ground.
Then the difference between the heights of the snow is:
3t- t = 2t
And we know that at 5:00 p.m. there were 12 inches of snow then we can solve the linear equation for t:
3t = 12in
t = (12in)/3 = 4 in
Replacing that in the difference of heights:
2t = 2*4in = 8in
From this, we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
If you want to learn more about linear equations:
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50. i’m sure you’re learning the proper mathematical way to do that, but my brain went ‘4+5 is 9. that’s 10 times less than 90’
Answer:
r=8 is a solution. Explanation: To tell whether the given value is a solution to the inequality, we have to solve the inequality. r−3≤9
r=8 is a solution. Explanation: To tell whether the given value is a solution to the inequality, we have to solve the inequality. r−3≤9 Add 3 ... More
Step-by-step explanation:
r=8 is a solution. Explanation: To tell whether the given value is a solution to the inequality, we have to solve the inequality. r−3≤9
r=8 is a solution. Explanation: To tell whether the given value is a solution to the inequality, we have to solve the inequality. r−3≤9 Add 3 ... Morep
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
