Answer:
The answer is 75 inches.
Step-by-step explanation:
If Alice is 10 inches, she is 2/3 as tall as the door. At her regular height, to be 2/3 as tall as the door the door would have to be 75 inches high.
I did this test b4, yours is answer #number 12
Convert things to their basic forms.
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>
<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>
<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>
<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>
<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>
<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>
<span>4. 1- cos^2θ </span>
<span>sin^2 </span>
<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>
<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>
<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>
<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>
<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>
<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>
<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>
<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
First you have to make it multiplication so you flip 5/2 around and the two “2’s” will cancel each other so it will be 17/5 or 3 2/5
Given: 
1. Combining like terms (here terms 6x and 2x both contain x, then we can combine them):
2. Distributive postulate:
The equation is
3. Zero product postulate (zero product postulate state that if a product of two factors is equal to zero, then first factor is equal to zero or second factor is equal to zero):
4. Subtraction property of equality:
a) subtract 6 from the first equation:
b) subtract 2 from the second equation:
Step-by-step explanation:
Given the linear equation, y = ⅔x + 1, where the <u>slope</u>, m = ⅔, and the y-intercept, (0, 1) where<em> b</em> = 1.
<h3><u>Start at the y-intercept:</u></h3>
In order to graph the given linear equation, start by plotting the coordinates of the y-intercept, (0, 1). As we know, the <u>y-intercept</u> is the point on the graph where it crosses the y-axis. It coordinates are (0, <em>b</em>), for which the value of b represents the value of the y-intercept in slope-intercept form, y = mx + b.
<h3><u>Plot other points using the slope:</u></h3>
From the y-intercept, (0, 1), we must use the slope, m = ⅔ (<em>rise</em> 2, <em>run</em> 3) to plot the other points on the graph. Continue the process until you have sufficient amount of plotted points on the graph that you could connect a line with.
Attached is a screenshot of the graphed linear equestion, which demonstrates how I plotted the other points on the graph using the "rise/run" techniques" discussed in the previous section of this post.
