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3 years ago

14

Cosxsinx-2cosx=0 I dont know how to solve this

1 answer:

jekas [21]3 years ago

8 0

Factor out cosx: cosx(sinx-2)=0
cosx=0 or sinx-2=0
cosx=0 or sinx=2
the largest value of sinx is 1, sinx will never be 2, so cosx=0, x=π/2 or 3π/2 are the two answers.

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78, 66, ,42,30 whats the missing number in the sequence

goldenfox [79]

Answer:

54

Step-by-step explanation:

30,42,54,66,78

counting by 12

4 0

3 years ago

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V^2+2=8-4v^2 solve the question algebically

iris [78.8K]

Answer:

v = ± sqrt(6/5)

Step-by-step explanation:

v^2+2=8-4v^2

Add 4v^2 to each side

v^2+ 4v^2+2=8-4v^2+ 4v^2

5v^2 +2 = 8

Subtract 2 from each side

5v^2 +2-2 = 8-2

5 v^2 = 6

Divide each side by 5

5v^2 /5 = 6/5

v^2 = 6/5

Take the square root of each side

sqrt(v^2) = ± sqrt(6/5)

v = ± sqrt(6/5)

8 0

3 years ago

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If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago

The length of a rectangular flower bed is 2ft longer than the width. If the area is 6ft, then what are the exact length and widt

Marizza181 [45]

Answer:

Exact dimensions:

$width=-1+\sqrt{7}$

$length=-1+\sqrt{7}+2$

$length=1+\sqrt{7}$

Approximate dimensions:

$width=1.64575ft$

$length=1.64575+2$

$length=3.64575ft$

Step-by-step explanation:

Let's assume width of rectangle is w ft

The length of a rectangular flower bed is 2ft longer than the width

so,

length =w+2

$L=w+2$

now, we can find area

$A=L\times W$

now, we can plug it

$A=(w+2)\times w$

$A=w^2+2w$

we are given area =6

so, we can set it equal

and then we can solve for w

$w^2+2w=6$

$w^2+2w-6=0$

we can use quadratic formula

$ax^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

now, we can compare and find a,b and c

a=1 , b=2 , c=-6

$w=\frac{-2\pm \sqrt{2^2-4\cdot \:1\left(-6\right)}}{2\cdot \:1}$

$w=-1+\sqrt{7},\:w=-1-\sqrt{7}$

we know that dimension can never be negative

so, we will only consider positive value

Exact dimensions:

$width=-1+\sqrt{7}$

$length=-1+\sqrt{7}+2$

$length=1+\sqrt{7}$

Approximate dimensions:

$width=1.64575ft$

$length=1.64575+2$

$length=3.64575ft$

7 0

3 years ago

PLS HELP ASAP<br><br> Solve the equation <br> 1/2x+3/4x=5-2.5x<br> X=_____

Trava [24]

Answer:

1 1/3

Step-by-step explanation:

x = 4/3 or 1 1/3

4 0

2 years ago

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