Answer:
Step-by-step explanation:
we are given
we have to find f/g
we can write as
now, we can plug values
and we get
so,
we know that denominator can not be zero
so,
now, we can solve for x
Divide both sides by 5
and we get
Math question shown below question and graph
1 answer:
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Answer:
big poop face looking butt with a massive forehead poopy duck dipers
Step-by-step explanation:
↑∴↔↔⇔⇔³∛÷≈≤≥≥β⇵⊕⊕¬⊕⊕⊕⊕⊕⊕¬∪∪⊥∡∠°∞㏒㏑∡Ф⇒⇒√³π∫↓∵∴∴ and thats why it is big poop face looking butt with a massive forehead poopy duck dipers
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)
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.
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,