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3 years ago

What’s 6cm and 2.5cm

Mathematics

2 answers:

forsale [732]3 years ago

6 0

Answer:

6cm + 2.5cm = 8.5cm :)

Gemiola [76]3 years ago

5 0

Answer:

Answer: 3.2

Step-by-step explanation:

: )

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If there are 5 terms in a quadratic sequence

exis [7]

Answer:

7.2

Step-by-step explanation:

First differences:

0.8-0.2, 1.8-0.8, 3.2-1.8, 5.0-3.2

0.6, 1.0, 1.4, 1.8

Second differences:

1.0-0.6, 1.4-1.0, 1.8-1.4

0.4, 0.4, 0.4

For the next term, add 0.4 to the last term from the first differences, and add the sum to the last term of the quadratic sequence:

0.4+1.8=2.2

2.2+5.0=7.2

∴ The next term in the sequence is 7.2

7 0

3 years ago

(200 x 3) + (50 x 3)

m_a_m_a [10]

200x3=600

50x3=150

600+150=750

750

6 0

3 years ago

Read 2 more answers

65/p • 3 when p = 5

Makovka662 [10]

Answer:

Step-by-step explanation:

65/5=13

13x3=39

8 0

3 years ago

Read 2 more answers

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: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$