All categories

3 years ago

Solve: 2cos^2(w)-cos(w)-1=0

Mathematics

2 answers:

netineya [11]3 years ago

7 0

Answer:

Step-by-step explanation:

2 cos²(w)-cos w-1=0

2 cos² w-2cos w+cos w-1=0

2 cos w(cos w-1)+(cos w-1)=0

(cos w-1)(2 cos w+1)=0

either cos w=1=cos 0=cos (0+2nπ)

w=2nπ,n∈I

or 2 cos w+1=0

cos w=-1/2=-cos (π/3)=cos ((2n+1)π±π/3)

w=[(2n+1)±1/3]π,

n∈ I

umka2103 [35]3 years ago

4 0

Answer:

w = π/2

Step-by-step explanation:

I used trig identities (double angle) to solve this problem.

Step 1: Rearrange problem

2cos²(w) - cos(w) - 1 = 0

2cos²(w) - 1 - cos(w) = 0

Step 2: Trig Identity

cos2w - cos(w) = 0

Step 3: Combine like terms

cosw = 0

Step 4: Solve

w = cos^-1(0)

w = π/2

I may have done it wrong so don't rely too much on me

You might be interested in

Help pplsllsls brainiest???

Anvisha [2.4K]

Hellor rnirde ejrjfndndodnff

5 0

3 years ago

Which expressions are equivalent to 4^{-2} \cdot 7^{-2}4 −2 ⋅7 −2 4, start superscript, minus, 2, end superscript, dot, 7, start

PolarNik [594]

Answer:

Step-by-step explanation:

Given the expression 4^{-2}•7^{-2}. The following expression are equivalent to given expression on simplification.

Generally from indices, a^-b = 1/a^b. Applying this to the given expression we have:

4^{-2}•7^{-2} = 1/4^2 • 1/7^2

= 1/(4×4) • 1/(7×7)

= 1/16 • 1/49

= 1/(16×49)

= 1/784

4 0

3 years ago

A rectangular parking lot has an area of 315 square yards. The width of the parking lot is 5 yards longer than the length of the

Blababa [14]

Answer:

Dimension of parking lot is approximately 15.425 × 20.425 yards.

Step-by-step explanation:

Given:

Area of the parking lot = $315 \ yd^2$

Let the length of the parking lot be $'l'$

Also Given:

The width of the parking lot is 5 yards longer than the length of the parking lot.

so we can say that;

width of the parking lot, $w=5+l$

We need to find the dimension of parking lot.

Solution:

Now we can say that;

Area of rectangle is equal to length times width.

framing in equation form we get;

$lw=315$

Now substituting the value of w in above equation we get;

$l(l+5)=315\\\\l^2+5l=315\\\\l^2+5l-315=0$

Now we will find the roots using quadratic equation.

$ax^2+bx+c =0$

a = 1

b=5

c=-315

Formula for quadratic equation is;

$l =\frac{-b\±\sqrt{b^2-4ac}}{2a}$

First we will find;

$\sqrt{b^2-4ac} = \sqrt{5^2-4\times1\times-315} =\sqrt{25+1260}=\sqrt{1285} \approx35.85$

Now we will substitute the value we get;

$l =\frac{-5\±35.85}{2\times1}= \frac{-5\±35.85}{2}$

Now we will find 2 roots we get;

$l= \frac{-5+35.85}{2}= \frac{30.85}{2}\approx15.425\\\\OR\\\\l =\frac{-5-35.85}{2}= \frac{-40.85}{2} \approx -20.425$

Now we get 2 values of length one is positive and other is negative.

But length of parking lot cannot be negative hence we will discard the negative value.

Hence the length of the parking lot = 15.425 yards

Width of the rectangular parking lot = $l+5=15.425+5 =20.425$

Hence we can say that dimension of parking lot is approximately 15.425 × 20.425 yards.

3 0

3 years ago

Need help on this question

miss Akunina [59]

If you divide the first equation by 2 and the second by 3, you get

$4x+3=y\\4x-1=y$

These equations differ only in their y-intercept, so describe parallel lines. The appropriate choice is ...

... H The lines are parallel.

7 0

3 years ago

Start at 5. Create a pattern that multiplies each number by 3 and subtract 2. Stop when you have five numbers.

RoseWind [281]

13, 37, 109, 325, 973. Hope that helps!

7 0

3 years ago