The equation that matches the given situation is $-2\frac{1}{2}+6\frac{3}{4}=?$ . So, after two moves, Eric's elevation changed $4\frac{1}{4}$ meters above.

We know that in mathematics, subtraction means removing something from a group or a number of things. What is left in the group gets smaller when we subtract. The minuend is the first element we use. The subtrahend is the part that is being removed. The difference is the portion that remains after subtraction.

Assume that "negative" means climbing down and "positive" means climbing up. We must locate the elevation change in this area.

Given that Eric climbed straight down $2\frac{1}{2}$ meters. So we can write $-2\frac{1}{2}$ .

Again Eric climbed straight up $6\frac{3}{4}$ meters. So, we can write $6\frac{3}{4}$ .

Then the change = $-2\frac{1}{2}+6\frac{3}{4}$

= $-\frac{(2*2+1)}{2}+\frac{6*4+3}{4}$

= $-\frac{5}{2} +\frac{27}{4}$

= $\frac{-(5*2)+27}{4}$

= $\frac{-10+27}{4}$

= $\frac{17}{4}$

= $\frac{4*4+1}{4}$

= $4\frac{1}{4}$

Therefore, the equation that matches the given situation is $-2\frac{1}{2}+6\frac{3}{4}=?$ . So, after two moves, Eric's elevation changed $4\frac{1}{4}$ meters above.

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1 year ago

Please help me with this question I’m so confused

zavuch27 [327]

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 10( \frac{1}{2} ) + \frac{1}{2} ( - 2) {( \frac{1}{2}) }^{2} \\ s = 10( \frac{1}{2} ) + \frac{1}{2} ( - 2)( \frac{1}{4} ) \\ s = 5 + ( - 1)( \frac{1}{4} ) \\ s = 5 + ( - \frac{1}{4} ) \\ s = 5 - \frac{1}{4} \\ s = \frac{20}{4} - \frac{1}{4} \\ s = \frac{19}{4}$