Meta

🔍

question:In triangle PAT, with angle P = 36^circ, angle A = 56^circ, and PA = 10, points U and G lie on sides TP and TA respectively, such that PU = AG = 1. If M and N are the midpoints of PA and UG, what is the measure of the acute angle formed by lines MN and PA when solved using complex numbers?

answer:The length 10 of side PA is not relevant in this problem, as it will cancel out. Our goal is to find the midpoint of the complex numbers representing text{cis }36^circ and text{cis }124^circ (where text{cis }x = cos x + isin x), and then determine the argument of this midpoint. Let p = -5 and a = 5 be the complex numbers corresponding to points P and A. The midpoint M of PA is at the origin, 0. Now, calculate the complex numbers for U, G, and N: 1. u = p + 1cdot (cos 36^circ + isin 36^circ) = -5 + text{cis }36^circ 2. g = a + 1cdot (cos 124^circ + isin 124^circ) = 5 + text{cis }124^circ 3. n = frac{1}{2}(u + g) To find n, we use the sum-to-product identities: n = cos 44^circ Big(text{cis }80^circBig) The argument of n corresponds to the angle formed by lines MN and PA. Hence, the measure of the acute angle is 80^circ.

question:List the vectors that form a basis for the null space of the following matrix, rounding your answer to three decimal places: left( begin{array}{ccccc} 1 & 8 & -9 & -3 & -3 -2 & -10 & 3 & -3 & 1 4 & 9 & -3 & 5 & 4 end{array} right)

answer:The basis for the null space of the matrix is given by the following vectors: left{ begin{array}{l} begin{bmatrix} -330.000 75.000 -19.000 0.000 147.000 end{bmatrix}, begin{bmatrix} -8.000 -2.000 -5.000 7.000 0.000 end{bmatrix} end{array} right}

question:What is the identity of the velvet-green growth with a flattened body and root-like structures that Norah found on a tree trunk?

answer:The growth on the tree trunk observed by Norah is likely a bryophyte, such as moss. Bryophytes are small, non-vascular plants that do not have true tissues. They often have a flattened body and attach to surfaces like tree trunks using root-like structures called rhizoids. Mosses are a well-known group of bryophytes, and they typically exhibit these characteristics.

question:Verify the identity: [ frac{1}{1+(n-1)x} - frac{1}{1+nx} = frac{x}{(1+(n-1)x)(1+nx)} ] for all integers ( n geq 1 ) and non-zero real numbers ( x ). Using this identity, prove that for ( x neq 0 ): [ sum_{n=1}^N frac{x}{(1+(n-1)x)(1+nx)}=frac{N}{1+Nx} ] Then, deduce that the infinite series [ sum_{n=1}^{infty} frac{1}{nleft(frac{3}{2}+nright)} ] is convergent and determine its sum to infinity.

answer:From the identity: [ frac{x}{(1+(n-1)x)(1+nx)}=frac{1}{1+(n-1)x}-frac{1}{1+nx} ] we can observe that for ( x > 0 ), the series [ sum_{n=1}^N left(frac{1}{1+(n-1)x}-frac{1}{1+nx}right) ] telescopes. To show convergence, it suffices to verify that [ lim_{Ntoinfty} frac{N}{1+Nx} ] exists and is finite for all ( x neq 0 ). The pattern in the series is revealed by examining the partial sums: [ begin{align*} n=1 &Rightarrow frac{x}{1+x} n=2 &Rightarrow frac{x}{1+x} + frac{x}{(1+x)(1+2x)} = frac{2}{1+2x} n=3 &Rightarrow frac{x}{1+x} + frac{x}{(1+x)(1+2x)} + frac{x}{(1+2x)(1+3x)} = frac{3}{1+3x} end{align*} ] By mathematical induction, it can be shown that for ( n = N ): [ sum_{k=1}^N frac{x}{(1+(k-1)x)(1+kx)} = frac{N}{1+Nx} ] This implies the infinite series [ sum_{n=1}^{infty} frac{1}{nleft(frac{3}{2}+nright)} ] converges to the limit: [ lim_{Ntoinfty} frac{N}{1+Ncdotfrac{3}{2}} = lim_{Ntoinfty} frac{2N}{3+2N} = frac{2}{3} ] Thus, the sum to infinity of the given infinite series is (frac{2}{3}).