Helppp meee please!!

Mathematics

2 answers:

MariettaO [177]3 years ago

6 0

Vertical stretch ununinin

kolbaska11 [484]3 years ago

3 0

Vertical stretch

Hope this helps! :)

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Can you pls help me with these questions?<br> I'm giving brainliest ^^

Eddi Din [679]

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A. 15, 9, 3, -3, -9

Step-by-step explanation:

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3 years ago

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Sever21 [200]

Answer:

Step-by-step explanation:

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3 years ago

Analyze the diagram below and complete the instructions that follow.

Trava [24]

Law of Cosines says

$b^2 = a^2 + c^2 - 2 ac \cos B$

$2 a c \cos B = a^2 + c^2 - b^2$

$\cos B = \dfrac{a^2 + c^2 - b^2}{2 a c}$

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$B = \arccos \dfrac{ 391}{512} \approx 40.2 ^\circ$

Choice C

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3 years ago

Find the length of the following two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 < t < 16

andrezito [222]

Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

$r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt$

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

$r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt$