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3 years ago

Pls help I’m not goof at simplifying

Mathematics

2 answers:

otez555 [7]3 years ago

8 0

It would be 4

5 - (- 1)
Would just be 5 - 1 which is 4

aivan3 [116]3 years ago

4 0

5-(-1) you need to change the negative one to a positive one
5+(+1)then just remove the parentheses and add them up witch gives
5+6

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Irina-Kira [14]

Answer:

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Step-by-step explanation:

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2 years ago

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4 years ago

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3 years ago

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Answer:

width: 1.9 ft
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The volume of a rectilinear shape is given by ...

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f(x) = x(x +0.4)(x -0.2)

Then we're looking for the value of x that satisfies ...

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3 years ago

Kaliska is jumping rope. The vertical height of the center of her rope off the ground R(t) (in cm) as a function of time t (in s

xz_007 [3.2K]

Answer:

R (t) = 60 - 60 cos (6t)

Step-by-step explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

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a + d = 0 ----- (1)

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i.e

$R ( \dfrac{\pi}{12}) = 60$

$60 = acos (\dfrac{b \pi}{12}) +d --- (2)$

Recall from the question that:

At t = 0, R(0) = 0 which is the minimum

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Similarly; 60 × 2 = maximum

R'(t) = -ab sin (bt) =0

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here;

k is the integer

making t the subject of the formula, we have:

$t = \dfrac{k \pi}{b}$

replacing the derived equation of k into R(t) = acos (bt) + d

$R (\dfrac{k \pi}{b}) = d+a cos (k \pi)$ $= \left \{ {{a+d \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.$

Since we known a < 0 (negative)

then d-a will be maximum

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a = -60 and d = 60

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$cos (\dfrac{\pi b}{12}) =0$

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∴

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7 0

3 years ago