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2 years ago

How do we solve this one

Mathematics

1 answer:

Whitepunk [10]2 years ago

7 0

Answer:

how should I answer it?

Step-by-step explanation:

elimination,substitution, etc

answer will be in comments

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Sixty less than ten times ten

Otrada [13]

Answer:

Step-by-step explanation:

10x10=100

100-60=40

Hope this helps! :)

3 0

3 years ago

How to write out one hundred four and 2thousands

lions [1.4K]

104.002 i think this is the answer

8 0

3 years ago

Read 2 more answers

Can you help find the inverse of the functions? (15 points)

anyanavicka [17]

hi so for the first exercise it’s the 3rd and for the second exercise it’s the first

Step-by-step When you have to find the inverses of numbers the sign ( -or+ ) never change see: the inverse of 2 is 1/2 or the inverse of (x+7)/(3x-5) is (3x-5)(x+7), hope you’ll understand : )

6 0

3 years ago

Use the Binomial Theorem to find the indicated coefficient or term.

Dmitriy789 [7]

Answer:

ans: coefficient of 3rd term is 60

3rd term is 60x

Step-by-step explanation:

(5x +2)³ = C(3,2) × (5x)^(3-2) × 2²

= 3 × 5 × 4 × x

= 60x

8 0

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

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0

2 years ago

How do we solve this one​

How do we solve this one