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3 years ago

8

What are the coordinates

= \sqrt{x - 3} " alt="y = \sqrt{x - 3} " align="absmiddle" class="latex-formula">
A(4;-1)
B(4;1)
C(3;1)
D(0;0)

1 answer:

SashulF [63]3 years ago

3 0

Answer: its A or b but i think A

Step-by-step explanation:

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zavuch27 [327]

Answer:

J'(2,-3), K'(4, -3) and L'(5,-2)

Step-by-step explanation:

The given triangle has vertices at J(2,3), K(4, 3) and L(5,2).

The transformation rule for a reflection in the x-axis is:

$(x,y)\to (x,-y)$

We substitute the points to obtain the coordinates of the image triangle J'K'L'

$J(2,3)\to J'(2,-3)$

$K(4,3)\to K'(4,-3)$

$L(5,2)\to L'(5,-2)$

In other words, we negate the y-coordinates of triangle JKL to obtain the coordinates of J'K'L'

See attachment for graph.

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3 years ago

Find f'(0.3) for f(x)= the integral from 0 to x of arccos(t)dt

Lady_Fox [76]

$\bf \textit{using the 2nd fundamental theorem of calculus}\\\\
\cfrac{dy}{dx}\displaystyle \left[ \int\limits_{0}^{x}\ cos^{-1}(t)dt \right]\implies cos^{-1}(x)
\\\\\\
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now.. 0.3 is just a value...we'e assuming Radians for the inverse cosine, so, if you check, make sure your calculator is in Radian mode

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Answer:

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Step-by-step explanation:

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3/5 = x/15

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