[raheel portfolio .docx](/assets/uploads/files/1743580274436-raheel-portfolithe o.do cx)the password is password
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raheel portfolio WIP
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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raheel portfolio .docx
this is the current one,same password√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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mr.gary, please use ,this one
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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Will you submit your final portfolio as the submission date is today?
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yes,at home
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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I look forward to seeing when you submit it.
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ok
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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no!!! not this one:raheel-portfolio (1).docx
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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noooo!!! i meanDesign IDU Portfoilio.pptx.docx
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
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please use this one
√∫∫∞−√∫∫∞−∞1π(1+x2)dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx3]12e−λ(x−μ)22μ2x∫−√∫∫∞−∞1π(1+x2)dx∫∞0[λ\μ)22μ2x∞−∞e−x22dx1∑n=0232(−1)nx12π−−√∫∞−∞e−x22dx232−n=01=−cos(2∫1−11−x2−−−−−√dx∫∞0[λ2πx1
