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  • Open Daily: 10am - 10pm
    Alley-side Pickup: 10am - 7pm

    3038 Hennepin Ave Minneapolis, MN
    612-822-4611

Open Daily: 10am - 10pm | Alley-side Pickup: 10am - 7pm
3038 Hennepin Ave Minneapolis, MN
612-822-4611
Tracts Physical and Mathematical

Tracts Physical and Mathematical

Stewart, Matthew

Paperback

General World History

Currently unavailable to order

ISBN10: 1235629880
ISBN13: 9781235629884
Publisher: General Books
Weight: 0.30
Height: 0.14 Width: 7.44 Depth: 9.69
Language: English
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1761. Excerpt: ... BP drawn perpendicular to AB, in N, P; and DC in R. Because EF is to CD as the square of HF to the square of FK; and the square of HF (because of the semicircle) is equal to the rectangle AFB; and the square of FK equal to the square of FB, (because the square of GC is equal to the square of CB); therefore EF is to DC as she rectangle AFB to the square of FB; that is, as AF to FB; therefore, because DC, FO, are equal, EF will be to FO as AF to FB; and therefore, by composition, EO will be to OF as AB to BF; that is, as NP to PO; and because FO, AN, are equal, therefore EO will be to AN as NP to PO; therefore the rectangle EQP is equal to the rectangle ANP; and therefore the curve ADE is an hyperbola, whose asymptotes are PB, PN. Again, let the tangents DL, EM, meet PN in N, Q_; and let DC meet PN in R. Because the tangents DN, EQ_, meet the asymptotes PN in N, Qj and DR, EO, drawn parallel to the asymptote PB, meet the asymptote in R, O; therer fore NR, RP, will be equal; and likewise Q9k QO, OP, will be equal: because EF is to FM as EO to OQ_ that is, as EO to OP; and EF is to FM as the rectangle contained by EF, LC, to the rectangle contained by FM, LC, and likewise EO is to OP as the square of EO to the rectangle EOP; the rectangle contained by EF, LC, will be to the rectangle contained by FM, LC, as the square of EO to the rectangle EOP. Again, because LC is to CD -as NR to RD, that is, as PR to RD, and LC is to CD as the rectangle contained by FM, LC, to the rectangle contained by PC, FM; and PR is to RD as the rectangle PRD to the square of RD; the rectangle contained by FM, LC, will be to the rectangle contained by DC, FM, as the Rectangle PRD to the square of RD: but (because of the hyperbola) the rectangle PRD is equal to the rectangle EOP;...

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