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3 years ago

7

Nate finishes his garden at 4:47 p. After pruning his roses for 18 minutes and watering his vegetables for 45 minutes. What time

does Nate finish gardening?

1 answer:

3241004551 [841]3 years ago

3 0

He finish his gardening at 5:50 p.m.

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What is the prime number of 40

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3 years ago

The grid shows Figure Q and its image Figure Q′ after a transformation.

Svetach [21]

The transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

<h3>What is transformation?</h3>

A transformation is a general term for four specific ways to manipulate the shape or position of a point, a line, or a geometric figure.

We have,

Coordinates of a pentagon Q.

(2, 4), (3, 7), (7, 5), (5,4), and (4,2).

Coordinates of pentagon Q'.

(-2, -4), (-3, -7), (-7, -5), (-5,-4), and (-4,-2).

We see that the coordinates of each point of the pentagon Q have changed their position as negative coordinates in pentagon Q'

The coordinates of pentagon Q are in (x, y) which can be assumed to be in the first quadrant of the coordinate plane.

The coordinates of pentagon Q' are in (-x, -y) which is in the 3rd quadrant.

From the first quadrant to 3rd quadrant there is a rotation of 180°.

Thus the transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

Learn more about transformation here:

brainly.com/question/4458799

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1 year ago

Answer fast and correctly.

IrinaVladis [17]

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3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

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3 years ago

Please help me ASAP!

Nookie1986 [14]

Answer: how do i edit iT?

Step-by-step explanation:

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