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For a square matrix of order n, multiplying every entry by a scalar k multiplies the determinant by k^n. Here n = 3, so |2A| = 2³|A| = 8 × 5 = 40.
The stroma is the fluid-filled space surrounding the thylakoid membranes, and it holds the enzymes such as RuBisCO that fix carbon dioxide. The thylakoid membranes host the light-dependent reactions instead.
Using the standard limit sin θ/θ → 1 as θ → 0, write (sin 3x)/x as 3 · (sin 3x)/(3x). The bracketed factor tends to 1, leaving the factor 3.
The favourable ordered pairs are (3,6), (4,5), (5,4) and (6,3), that is 4 of the 36 equally likely outcomes. So the probability is 4/36, which simplifies to 1/9.
Differentiating gives f′(x) = 3x² − 3 = 3(x − 1)(x + 1), so the stationary points are x = ±1. Since f″(x) = 6x, we get f″(−1) = −6 < 0, confirming a local maximum at x = −1.
The pulmonary veins return oxygenated blood from the lungs to the left atrium, making them the usual exception to the rule that veins carry deoxygenated blood. The pulmonary artery instead carries deoxygenated blood from the right ventricle to the lungs.
EcoRI recognises the palindromic hexanucleotide 5′-GAATTC-3′ and cuts between G and A, generating complementary sticky ends. The other sequences listed are recognition sites of different restriction enzymes.
Perpendicular vectors have zero dot product, so (2)(1) + (3)(−λ) + (1)(2) = 4 − 3λ = 0. Solving gives λ = 4/3.
Let the integral be I and substitute x → −x; the integrand becomes eˣ cos x/(1 + eˣ). Adding the two expressions gives 2I = ∫ cos x dx over [−π/2, π/2] = 2, hence I = 1.
Dividing throughout by 144 gives x²/16 − y²/9 = 1, so a² = 16 and b² = 9. For a hyperbola e = √(1 + b²/a²) = √(1 + 9/16) = √(25/16) = 5/4.
The conjugate is z̄ = 3 + 4i, so z z̄ = (3 − 4i)(3 + 4i) = 9 + 16 = 25. This product equals the square of the modulus, not the modulus itself.
The electron transport chain complexes are embedded in the inner mitochondrial membrane. The mitochondrial matrix contains the enzymes of the Krebs cycle, while the outer membrane is freely permeable to many small molecules.
Squaring sin θ + cos θ gives 1 + 2 sin θ cos θ = 1/4. Therefore 2 sin θ cos θ = −3/4, so sin θ cos θ = −3/8.
Abscisic acid is the stress hormone that signals guard cells to lose turgor and close stomata. Auxins and gibberellins mainly promote growth, while cytokinins promote cell division.
An idempotent matrix satisfies A² = A, so its eigenvalues can only be 0 or 1. If both eigenvalues were 1, the matrix would be the identity, which is excluded, so at least one eigenvalue is 0 and det A = 0.
The lac repressor is encoded by lacI and binds the operator to block RNA polymerase access to the promoter. Lactose, as allolactose, relieves repression by binding the repressor.
The first ball is red with probability 5/8. Given that the first is red, the second is red with probability 4/7, so the required probability is (5/8)(4/7) = 20/56 = 5/14.
Photosystem II replaces its lost reaction-center electrons by extracting electrons from water through the oxygen-evolving complex. This water splitting releases oxygen and protons and supplies electrons to the photosynthetic electron transport chain.
Let I = ∫₀^{π/2} ln(tan x) dx. Substituting x → π/2 − x gives I = ∫₀^{π/2} ln(cot x) dx = −I, because ln(cot x) = −ln(tan x); hence 2I = 0 and I = 0.
The first heart sound, S1, is caused by closure of the atrioventricular valves at the start of ventricular systole. Closure of the semilunar valves produces the second heart sound.
For independent events, P(A ∩ B) = P(A)P(B). Here 0.4 × 0.5 = 0.20, so the joint probability is 0.20.
Lysosomes are membrane-bound organelles containing hydrolytic enzymes that digest macromolecules and worn-out cell components. This digestive role explains the historical nickname 'suicidal bag'.
The derivative of e^{u} is e^{u} times du/dx. With u = 3x, du/dx = 3, so dy/dx = 3e^{3x}.
The curves intersect where x² = 2x, giving x = 0 and x = 2. Between these points 2x lies above x², so area = ∫ from 0 to 2 of (2x − x²) dx = [x² − x³/3] from 0 to 2 = 4 − 8/3 = 4/3.
A heterozygous tall plant (Tt) produces T and t gametes, while the dwarf parent (tt) produces only t. Offspring are Tt (tall) and tt (dwarf) in equal numbers, so the ratio is 1 tall : 1 dwarf.
The volume equals |a · (b × c)|, the absolute value of the determinant with the three vectors as rows. The determinant is 1(1 − 0) − 1(0 − 1) + 0 = 2, so the volume is 2 cubic units.
During DNA replication, helicase breaks the hydrogen bonds between complementary bases and unwinds the double helix ahead of the replication fork. The other enzymes listed act after or during synthesis rather than performing this unwinding step.
Most glucose is reabsorbed from the filtrate in the proximal convoluted tubule by secondary active transport with sodium. This segment has abundant microvilli and transporters for efficient reabsorption.
By Vieta's formulas, α + β = 6 and αβ = 4. Using α³ + β³ = (α + β)³ − 3αβ(α + β) gives 216 − 3 × 4 × 6 = 216 − 72 = 144.
A woman with group AB has genotype IAIB, and a group O man has genotype ii. Their children receive i from the father and either IA or IB from the mother, giving IAi (group A) or IBi (group B) in equal numbers.
Rewrite the expression as (5/3)·sin(5x)/(5x). As x→0, sin(5x)/(5x)→1, so the limit is 5/3. The denominator 3 remains in the coefficient and must not be discarded.
The magnitude of the cross product is |a||b| sin θ. Substituting gives 3 × 4 × sin 30° = 12 × 1/2 = 6. The angle used is the angle between the vectors, not its complement.
ATP hydrolysis releases free energy that can be coupled directly to many cellular reactions. Glucose is a fuel store, while NADH mainly carries electrons to the electron transport chain. GTP is used in some specific processes but is not the general energy currency.
The sum-8 outcomes are (2,6), (3,5), (4,4), (5,3), and (6,2), giving five equally likely cases. Of these, (3,5) and (5,3) contain a 5, so the conditional probability is 2/5.
Multiply numerator and denominator by the conjugate 1 + i. This gives (1 + i)²/(1² + 1²) = 2i/2 = i, so z = i. Therefore z⁴ = i⁴ = 1.
In C4 plants, PEP carboxylase fixes CO2 onto phosphoenolpyruvate to form oxaloacetic acid, the first stable four-carbon product. Ribulose bisphosphate is the CO2 acceptor in the Calvin cycle of C3 plants, not the primary mesophyll acceptor in C4 plants.
The corpus luteum secretes progesterone, which maintains the endometrium and supports early pregnancy. Luteinizing hormone triggers ovulation, while follicle-stimulating hormone stimulates follicle development.
If λ is an eigenvalue of A, then p(λ) = λ² − 2λ + 1 is an eigenvalue of p(A). For λ = 1, 2, 3, these values are 0, 1 and 4, so the determinant is 0 × 1 × 4 = 0.
A human somatic cell has 46 chromosomes. After DNA replication in S phase, each chromosome consists of two sister chromatids, so the G2 cell contains 46 × 2 = 92 DNA molecules.
Using integration by parts with u = x and dv = e^(−x) dx gives [−x e^(−x)]₀¹ + ∫₀¹ e^(−x) dx. This evaluates to −1/e + (1 − 1/e) = 1 − 2/e.
Differentiate using the chain rule: f′(x) = 2x cos(x²). Substituting x = 0 gives f′(0) = 2(0)cos(0) = 0, since cos(0) = 1.
The stroma is the fluid-filled space surrounding the thylakoids and contains the enzymes of the Calvin cycle. The thylakoid membranes house the light-dependent electron transport machinery.
For an n × n matrix, multiplying the entire matrix by a scalar k multiplies the determinant by kⁿ. Here n = 2 and k = 2, so det(2A) = 2² det(A) = 4 × 5 = 20.
Crossing over occurs during prophase I of meiosis, when homologous chromosomes pair as bivalents and non-sister chromatids exchange segments. This recombination increases genetic variation in the resulting gametes.
VITEEE has six letters, with E repeated three times. The number of distinct permutations is 6!/3! = 720/6 = 120.
Carbonic anhydrase catalyzes CO₂ + H₂O ⇌ H₂CO₃, so the immediate product is carbonic acid. Carbonic acid can then dissociate into bicarbonate and hydrogen ions.
For direction angles α, β, γ, the identity cos²α + cos²β + cos²γ = 1 holds. Thus cos²γ = 1 − 0 − 1/4 = 3/4, giving γ = 30° or 150°; the acute angle is 30°.
Colour blindness is X-linked recessive. The woman must be a carrier because her father was affected, so she is X^C X; the father is X Y. A son receives Y from his father and either X or X^C from his mother, giving a 1/2 chance of being colour-blind.
Write sin³x = sin x(1 − cos²x) and substitute u = cos x. The integral becomes ∫₀¹(1 − u²)du = [u − u³/3]₀¹ = 1 − 1/3 = 2/3.
The polynomial factors as (x − 2)²(x + 1). Thus the roots are x = 2, which has multiplicity two, and x = −1. There are therefore two distinct real roots.