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2 years ago

Ab-2/3-67*4 if a=23 and b=3

Mathematics

1 answer:

Viefleur [7K]2 years ago

7 0

what upAnswer:

Step-by-step explanation:

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If the equation y = 1/3 x + 2 is changed to y = 3x + 2, what will be true about the graph of the equation?

Anna11 [10]

That would be D) The slope of the graph will be changed.
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3 years ago

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Step-by-step explanation:

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3 years ago

If $x$ is the average of $13$, $-16$, and $6$ and if $y$ is the cube root of $8$, find $x^2 + y^3$. Please Help

Alekssandra [29.7K]

Answer:

If we have a set of N elements:

{x₁, x₂, ..., xₙ}

The mean value (also called the average value) Is calculated as:

Average = (x₁ + x₂ + ... + xₙ)/n

So if x is the average of 13, -16 and 6 ( a total of 3 values)

x will be equal to:

x = (13 + (-16) + 6)/3 = (19 - 16)/3 = 3/3 = 1

x = 1

And we know that:

y = ∛8

Remember that:

2*2 = 4

and

4*2 = 8

then

2*2*2 = 2^3 = 8

then ∛8 = 2.

So we have:

y = ∛8 = 2

Now we can replace these values in the equation:

x^2 + y^3

replacing:

x = 1

y = 2

we get:

1^2 + 2^3 = 1 + 8 = 9

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