# Ax Sentence Examples

- He swung the
**ax**once more, now only half a body length away. - Again she swung the
**ax**, and this time it went half way through the log. - He swung the
**ax**again. - She relinquished the
**ax**for the glass of tea. - "That's called an ice
**ax**, or piolet," Ryland answered. - Dean leaped on Shipton, clawing away at the soft snow, pummeling him like an eighth grade schoolyard brawler while Shipton, still clutching his ice
**ax**in one hand, swung at Dean, catching him on the cheek and face with the side of the solid handle. - Shipton flailed out at him with his
**ax**, missing his head by inches as Dean leaned sideways and frantically fumbled with his line to drop again. - The leg wound from Shipton's flailing ice
**ax**had been an eight-stitcher of no permanent consequence. - L
**ax**2 2**ax**i l This is called the kth transvectant of f over 4); it may be conveniently denoted by (f, (15)k. - I suppose you've got a good reason why you tried to beat the brains out of a guy holding an ice
**ax**, in the middle of the street with a bunch of people watching. - He swung his ice
**ax**into the wall in front of him, dug in the toes of his crampons and began to ascend toward Dean. - Shipton swung his ice
**ax**again, inching up closer to Dean. - Shipton's
**ax**bit the ice scarcely a foot below Dean as the man glared up at him, a snarl on his face. - His muscular back glistened with perspiration as he swung the
**ax**, expertly splitting a chunk of wood. - This time the
**ax**sank about four inches into the wood - in another spot. - After a full minute of tugging and grunting she managed to dislodge the
**ax**from the wood. **Ax**., Green**axis**creeping on the surface of damp soil; rh., colorless rhizoids penetrating the soil; asc.**ax**., ascendingof green cells.**axes**- Such curves are given by the equation x 2 - y 2 =
**ax**4 -1bx 2 y 2 +cy 4 . - In the theory of surfaces we transform from one set of three rectangular
to another by the substitutions 'X=' by+ cz, Y = a'x + b'y + c'z, Z =a"x+b"y-l-c"z, where X 2+Y2+Z2 = x2+ y2+z2.**axes** - - 2 ay2'
**ax**, Ux2 ï¿½ï¿½ï¿½ a y n ay. - Taking two of the equations
**ax**+ +cz) x"' 1 +... - It may be written in the form n n-1 2
**ax**1 +bx1 x2 +cx 1 x 2 + ...; or in the form n n n=1 n n-2 2 +(1)bx x2+ ? - Other forms are n-1 n-2 2
**ax**+nbx x +n(n-i)cx x +..., 1121 2 the binomial coefficients C) being replaced by s!(e), and n 1, n-1 1 n-2 2**ax**1 +l i ox l 'x 2 + L ?cx 1 'x2+..., the special convenience of which will appear later. - As between the original and transformed quantic we have the umbral relations A1 = A1a1 d-A2a2, A2 = /21a1+/22a2, and for a second form B1 =A 1 b 1+ A 2 b 2, B 2 =/21bl +ï¿½2b2ï¿½ The original forms are
**ax**, bi, and we may regard them either as different forms or as equivalent representations of the same form. - From the three equations
**ax**= alxl+ a2x2, b.= blxl+b2x2, cx = clxi+c2x2, we find by eliminating x, and x 2 the relation a x (bc)+b x (ca) +c x (ab) =0. - 32
**ax**l ay 2 ax2ay1' which, operating upon any polar, causes it to vanish. - The two forms
**ax**, bx, or of, 0, may be identical; we then have the kth transvectant of a form over itself which may, or may not, vanish identically; and, in the latter case, is a covariant of the single form. - The process of transvection is connected with the operations 12; for?k (a m b n) = (ab)kam-kbn-k, (x y x y or S 2 k (a x by) x = 4))k; so also is the polar process, for since f k m-k k k n - k k y = a x by, 4)y = bx by, if we take the k th transvectant of f i x; over 4 k, regarding y,, y 2 as the variables, (f k, 4)y) k (ab) ka x -kb k (f, 15)k; or the k th transvectant of the k th polars, in regard to y, is equal to the kth transvectant of the forms. Moreover, the kth transvectant (ab) k a m-k b: -k is derivable from the kth polar of
**ax**, viz. - Since, If F = An, 4) = By, 1 = I (Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn
**Ax**I**Ax**2**Axe**Ax1) J The First Transvectant Differs But By A Numerical Factor From The Jacobian Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, ï¿½) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (**Axby**A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1. - It is (f = (ab) 2 a n-2 r7 2 =Hx - =H; unsymbolically bolically it is a numerical multiple of the determinant a2 f a2f (32 f) 2ï¿½ It is also the first transvectant of the differxi
**ax****axa**x 2 ential coefficients of the form with regard to the variables, viz. - In general for a form in n variables the Hessian is 3 2 f 3 2 f a2f
**ax**i**ax**n**ax**2**ax**" ï¿½ï¿½ '**axn**and there is a remarkable theorem which states that if H =o and n=2, 3, or 4 the original form can be exhibited as a form in I, 2, 3 variables respectively. - If the forms be
**ax**, b2, cy,... - Moreover the second term on the left contains (a f)' c -2b z 2 = 2 (a f) k-2b x 2 - (b) /0-2a 2 ï¿½ if k be uneven, and (af)?'bx (i f) of) '-la if k be even; in either case the factor (af) bx - (bf)
**ax**= (ab) f, and therefore (n-k),bk+1 +Mï¿½f = k(n-2)f.(uf)uxn-2k-1; and 4 ' +1 is seen to be of the form f .14+1. - Y1 = x 15+f2n; fï¿½ y2 =x2-f?n, f .a b =
**ax**+ (a f) n, l; n u 2 " 2 22 2 +` n) u3 n-3n3+...+U 2jnï¿½ 3 n Now a covariant of**ax**=f is obtained from the similar covariant of ab by writing therein x i, x 2, for yl, y2, and, since y?, Y2 have been linearly transformed to and n, it is merely necessary to form the covariants in respect of the form (u1E+u2n) n, and then division, by the proper power of f, gives the covariant in question as a function of f, u0 = I, u2, u3,...un. - The Binary Quadratic.-The complete system consists of the form itself,
**ax**, and the discriminant, which is the second transvectant of the form upon itself, viz.: (f, f') 2 = (ab) 2; or, in real coefficients, 2(a 0 a 2 a 2 1). - =y la xl -i-y2a x2 must also vanish for the root a, and thence
**ax**, and a must also vanish for the same root; which proves that a is a double root of f, and f therefore a perfect square. - If a, a be the linear forms, above defined, he raises the identity
**ax**(0) =**ax**(aJ3) - (3x(aa) to the fifth power (and in general to the power n) obtaining (aa) 5 f = (a13) 5 az - 5 (a0) 4 (aa)**ax**?3 -F... - When C vanishes j has the form j = pxg x, and (f,j) 3 = (ap) 2 (aq)
**ax**= o. - Hence, from the identity
**ax**(pq) = px (aq) -qx (ap), we obtain (pet' = (aq) 5px - 5 (ap) (aq) 4 pxg x - (ap) 5 gi, the required canonical form. - When a z and the invariants B and C all vanish, either A or j must vanish; in the former case j is a perfect cube, its Hessian vanishing, and further f contains j as a factor; in the latter case, if p x,
**ax**be the linear factors of i, f can be expressed as (pa) 5 f =cip2+c2ay; if both A and j vanish i also vanishes identically, and so also does f. - Thus what have been called seminvariants are not all of them invariants for the general substitution, but are invariants for the particular substitution xl = X11 + J-s12, X 2 = 112 Again, in plane geometry, the most general equations of substitution which change from old
inclined at w to new**axes**inclined at w' =13 - a, and inclined at angles a, l3 to the old**axes****axis**of x, without change of origin, are x-sin(wa)X+sin(w -/3)Y sin w sin ' _sin**ax**y sin w a transformation of modulus sin w' sin w' The theory of invariants originated in the discussion, by George Boole, of this system so important in geometry. - Then of course (AB) = (ab) the fundamental fact which appertains to the theory of the general linear substitution; now here we have additional and equally fundamental facts; for since A i = Xa i +,ia2, A2= - ï¿½ay + X a2, AA =A?-}-A2= (X2 +M 2)(a i+ a z) =aa; A B =AjBi+A2B2= (X2 +, U2)(albi+a2b2) =ab; (XA) = X i A2 - X2 Ai = (
**Ax**i + /-Lx2) (- /-jai + Xa2) - (- / J.x i '+'**Axe**) (X a i +%Ga^2) = (X2 +, u 2) (x a - = showing that, in the present theory, a a, a b, and (xa) possess the invariant property. - (ab), aa, ab, (xa),
**ax**, xx. - To assist us in handling the symbolic products we have not only the identity (ab) cx + (bc) a x + (ca) bx =0, but also (ab) x x+ (b x) a + (
**ax**) b x = 0, (ab)a+(bc)a s +(ca)a b = 0, and many others which may be derived from these in the manner which will be familiar to students of the works of Aronhold, Clebsch and Gordan. - There is no linear covariant, since it is impossible to form a symbolic product which will contain x once and at the same time appertain to a quadratic. (v.) is the Jacobian; geometrically it denotes the bisectors of the angles between the lines
**ax**, or, as we may say, the common harmonic conjugates of the lines and the lines x x . - He was armed with a musketoon (which he carried rather as a joke), a pike and an
**ax**, which latter he used as a wolf uses its teeth, with equal ease picking fleas out of its fur or crunching thick bones. - Tikhon with equal accuracy would split logs with blows at arm's length, or holding the head of the
**ax**would cut thin little pegs or carve spoons. - He had a musketoon over his shoulder and an
**ax**stuck in his girdle. - So I went for them with my
**ax**, this way: 'What are you up to?' says I. - But his cry came an instant too late as Shipton plummeted past him, his ice
**ax**swinging in a rip across Dean's calf as he plummeted backward into space, and down to the rocks and churning river below. - She gripped the
**ax**handle. - Taking a few steps back she gripped the
**ax**half way down on the handle and slammed it down against the block of wood with a dull whack. - The
**ax**blade went about an inch into the wood. - He lifted the
**ax**, taking aim at a new block of wood. - So they filled a small boat with the things that he would need the most--an
**ax**, a hoe, a kettle, and some other things. - An
**ax**will be useful, a hunting spear not bad, but a three-pronged fork will be best of all: a Frenchman is no heavier than a sheaf of rye. - He hefted the
**ax**to his shoulder and grinned down at her. - The ï¿½th polar of
**ax**with regard to y is n-ï¿½ a aye i.e. - The complete covariant and contravariant system includes no fewer than 34 forms; from its complexity it is desirable to consider the cubic in a simple canonical form; that chosen by Cayley was
**ax**3 +by 3 + cz 3 + 6dxyz (Amer. - Hit him with an
**ax**, eh!...