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2 years ago

A bakery celebrated its 25th anniversary

Mathematics

1 answer:

Scilla [17]2 years ago

4 0

Answer:

Step-by-step explanation:

9: 9, 18, 27, 36, 45, 54, 63, 72

12: 12, 24, 36, 48, 60, 72

72 is the first number that can be divided (evenly) by both 9 and 12

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Help pls! Giving brainliest!!!!!

RUDIKE [14]

Answer:

$27 = 3 \times 3 \times 3\\\\\frac{1}{27} = \frac{1}{3 \times 3 \times 3} = \frac{1}{3^{3}} = 3^{-3}$

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2 years ago

FREE 10 POINTSS Which rule would translate triangle ABC 5 units to the left and 6 units up?

Elena-2011 [213]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Which are the solutions of x² = 19x + 12

Sever21 [200]

Answer:

$\{x=\frac{19-\sqrt{409} }{2}\ , \ x=\frac{19+\sqrt{409} }{2}\}$

Step-by-step explanation:

x² = 19x + 12

⇔ x² - 19x - 12 = 0

Calculating the discriminant :

b² - 4ac = (-19)² - 4×1×(-12) = 409

The discriminant is positive ,then the equation has two solutions.

$x=\frac{19-\sqrt{409} }{2} \ \ or\ \ x=\frac{19+\sqrt{409} }{2}$

4 0

2 years ago

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$