Answer:
I dont know what you are supposed to answer with but I'll try to help you out. I take it that you have to find the letter ex. the letter <em>d</em> in 1/<em>2d</em> = 7. What you have to do is to get the <em>d</em> by itself and in order to do that you must move the 1/2 over to the 7. So what you would have to do is take the fraction that has the letter on it and move it to the whole number side or the side without the letter but without moving the <em>d</em><em>.</em><em> </em>The <em>d</em> stays on one side while the other numbers are on the other side. When you move the 1/2 over it should look like this sort of due to the limitations of only being able to use text: <em>d</em> = 7 ÷ 1/2. Now after that you have to divide 7 by 1/2 or .5 for question 13 in this instance. Your answer for 13 should be this: <em>d</em><em> </em><em>=</em><em> </em><em>4.5</em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Make</em><em> </em><em>sure</em><em> </em><em>to</em><em> </em><em>show</em><em> </em><em>all</em><em> </em><em>your</em><em> </em><em>work</em><em> </em><em>just</em><em> </em><em>in</em><em> </em><em>case.</em>
13. <em>d</em> = 4.5
14. 18 = <em>f</em>
<em>1</em><em>5</em><em>.</em><em> </em><em>-</em><em>9</em><em> </em>=<em> </em><em>s</em>
<em>1</em><em>6</em><em>.</em><em> </em>-24 = <em>r</em>
17. 0.125 = <em>y</em>
18. -3 = <em>v</em>
Hope I could help as I am also a fellow 9th grader
Answer:
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For this case we have that by definition of trigonometric relations of rectangular triangles, that the sine of an angle is given by the opposite leg to the angle on the hypotenuse of the triangle. So:
Where h is the hypotenuse.
We cleared h:
We rationalize:
ANswer:
Option A
The answer is (-1,7)
I got this answer because to solve a system, you find the point on the graph where both lines intersect. And that point in this graph is (-1,7). Hope this helps!:)))))
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
