Answer:
-x^5 - 7x^4 + x^2 + (1/8)x - 9
Step-by-step explanation:
Rewrite the given -9+1/8x-7x^4+x^2-x^5 in descending order of the variable x:
-9+1/8x
-x^5 - 7x^4 + x^2 + (1/8)x - 9
 
        
             
        
        
        
The slope intercept form is y=mx+b, and b represents the y-intercept. In the given equation C=45t+100, 100 is the y-intercept. It represents the one-time fee to join the gym membership.
 
        
             
        
        
        
Check the picture below.
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AA%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B%5B6-%28-4%29%5D%5E2%2B%5B8-%28-2%29%5D%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B%286%2B4%29%5E2%2B%288%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B10%5E2%2B10%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B10%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BAC%3D10%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AB%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%0AD%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%5Cqquad%20%5Cqquad%20BD%3D%5Csqrt%7B%5B4-%28-2%29%5D%5E2%2B%5B0-6%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B%284%2B2%29%5E2%2B%28-6%29%5E2%7D%5Cimplies%20BD%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B6%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BBD%3D6%5Csqrt%7B2%7D%7D)
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,
![\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20four%20congruent%20triangles%7D%7D%7B4%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%283%5Csqrt%7B2%7D%29%285%5Csqrt%7B2%7D%29%20%5Cright%5D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%20%28%5Csqrt%7B2%7D%29%5E2%29%20%5Cright%5D%7D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%202%29%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A4%5B15%5D%5Cimplies%2060) 
 
        
        
        
Answer:
The answer is 5x^2+32x-21
Step-by-step explanation:
To find the area of a rectangle, you have to multiply the length by the width. Then, you will get the equation of (5x-3)(x+7). Then, multiply these two together, therefore getting 5x^2+32x-21.
 
        
             
        
        
        
The answer to this question is -3/35