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3 years ago

10

What is a correct name for the angle shown?

2 answers:

Masteriza [31]3 years ago

4 0

Answer:

the answer is option B. angle S.

when naming an angle we place the vertex of the angle in the middle. here the angle is RST. But that option is unavailable. very often when there are no other angles interfering with the parent angle, we represent it using one letter that is the mid letter, the vertex. here in this case it is S.

andreyandreev [35.5K]3 years ago

4 0

It would be S. the others don’t make sense.

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Step-by-step explanation:

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Read 2 more answers

Find dy/dx by implicit differentiation.

kow [346]

dy/dx by implicit differentiation is cos(πx)/sin(πy)

<h3>How to find dy/dx by implicit differentiation?</h3>

Since we have the equation

(sin(πx) + cos(πy)⁸ = 17, to find dy/dx, we differentiate implicitly.

So, [(sin(πx) + cos(πy)⁸ = 17]

d[(sin(πx) + cos(πy)⁸]/dx = d17/dx

d[(sin(πx) + cos(πy)⁸]/dx = 0

Let sin(πx) + cos(πy) = u

So, du⁸/dx = 0

du⁸/du × du/dx = 0

Since,

du⁸/du = 8u⁷ and

du/dx = d[sin(πx) + cos(πy)]/dx

= dsin(πx)/dx + dcos(πy)/dx

= dsin(πx)/dx + (dcos(πy)/dy × dy/dx)

= πcos(πx) - πsin(πy) × dy/dx

So, du⁸/dx = 0

du⁸/du × du/dx = 0

8u⁷ × [ πcos(πx) - πsin(πy) × dy/dx] = 0

8[(sin(πx) + cos(πy)]⁷ × (πcos(πx) - πsin(πy) × dy/dx) = 0

Since 8[(sin(πx) + cos(πy)]⁷ ≠ 0

(πcos(πx) - πsin(πy) × dy/dx) = 0

πcos(πx) = πsin(πy) × dy/dx

dy/dx = πcos(πx)/πsin(πy)

dy/dx = cos(πx)/sin(πy)

So, dy/dx by implicit differentiation is cos(πx)/sin(πy)

Learn more about implicit differentiation here:

brainly.com/question/25081524

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2 years ago

Please tell me the answer quick

fomenos

Answer:

D) The expression 9 - p has exactly 2 terms

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3 years ago

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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3 years ago