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2 years ago

Pls help ill give brainliest

Mathematics

1 answer:

ki77a [65]2 years ago

7 0

The answer would be:

3q x (2q + 1) x (q^2 - 4)

Hope this helped :)

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An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu

exis [7]

Take the derivative with respect to t
$- w \sin(wt) + \sqrt{8} w cos(wt)$
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
$0 = -w \sin(wt) + \sqrt{8} w cos(wt)$
divide by w
$0 =- \sin(wt) + \sqrt{8} cos(wt)$
we add sin(wt) to both sides

$\sin(wt)= \sqrt{8} cos(wt)$
divide both sides by cos(wt)
$\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}$ OR $\\$ $wt=arctan( { \frac{1}{ \sqrt{2} } )$
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

$I=cos(2n \pi -2arctan( \sqrt{2} ))$
since 2npi is just the period of cos
$cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 
$
substituting our second soultion we get
$I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))$
since 2npi is the period
$I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}$
so the maximum value = $\frac{1}{3}$
minimum value = $- \frac{1}{3}$

4 0

2 years ago

A rectangle has a length of 12 yards less than 3 times its width. If the area of the rectangle is 135 square yards, find the len

Aleonysh [2.5K]

Answer:

The length is 15 yards

Step-by-step explanation:

Given

Represent the width with W and length with L: So

$Area = 135yd^2$

$L=3W - 12$

Required

Determine the length of the rectangle:

Area is calculated as thus:

$Area = L * W$

Substitute 3W - 12 for L and 135 for Area

$135= (3W - 12) * W$

$135= 3W^2 - 12W$

Reorder:

$3W^2 - 12W - 135= 0$

Divide through by 3

$W^2 - 4W - 45 = 0$

Expand:

$W^2 - 9W + 5W - 45 = 0$

$W(W - 9) + 5(W - 9) = 0$

$(W + 5)(W-9) = 0$

$W + 5 = 0$ or $W - 9 = 0$

$W = -5$ or $W = 9$

But width can't be negative

So:

$W = 9$

Recall that: $L=3W - 12$

$L = 3 * 9 - 12$

$L = 15$

Hence, the length 15 yards

4 0

3 years ago

1/2(2+a)=3a+4/3 solve it

Mumz [18]

1/2(2+a)=3a+4/3

first, you use the distributive property

then your problem changes to...

1 + 1/2a = 3a + 4/3

then you subtract 1/2a with 3a

1 = 2 1/2a +4/3

now you subtract 1 with 4/3

-1/3 = 2 1/2a

now you divide -1/3 with 2 1/2a

-2/15 = a

A = -2/15 is your answer. Hope this helps :)

4 0

3 years ago

Solve the system of inequalities to decide if the point (-3,2) is part of the solution; y<=-4x-3, +8y>=7

inna [77]

Answer:

$(x,y) = (-3,2)$ is a solution

Step-by-step explanation:

Given

$y \le -4x + 3$

$x + 8y \ge 7$

Required

Determine if $(x,y) = (-3,2)$ is a solution

$y \le -4x + 3$ becomes

$2 \le -4 * -3 + 3$

$2 \le 15$ --- this is true

$x + 8y \ge 7$

$-3+ 8 * 2 \ge 7$

$13 \ge 7$ --- this is also true

Hence,

$(x,y) = (-3,2)$ is a solution

6 0

2 years ago

Please help me with this!! Number 6 is the problem

artcher [175]

The first two consecutive odd integers are 1 and 3.
The height has to be 3 since the question implies that the height is exceeds the base. So the base is 1. b = 1 cm

The area of a triangle is base*height*(1/2)
Plug in the values for base and height.
1*3*(1/2)

Simplify. Your final answer should be 1.5 cm^2

8 0

3 years ago