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3 years ago

A bedroom has a length of x + 6 feet and a width of x feet. Find the area of the bedroom, y, when x = 12.

Mathematics

2 answers:

Juliette [100K]3 years ago

8 0

The answer is 216 because (12 + 6) x (12)

Mumz [18]3 years ago

5 0

Step-by-step explanation:

A=l×b

y=x(x+6)

y=x^2+6x

when x is 12

y=12^2+6(12)

y=144+72

y=216 ft

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225% of what number is 81

Airida [17]

To find a percentage of a number we multiply the percentage times the number and get the result, here we have the percentage and the result and need to find out which number it corresponds to, lets write an equation to do so and keep in mind that the percentage itself can be written in a different way:
225% = 225 out of 100 = 225/100
So we have this equation, where we call x the number we are looking for:
<span>(225%)(x) = 81
</span>(225/100)(x) = 81
so we solve for x:
225x = 81*100 = 8100
x = 8100/225
x = 36
The original number is 36, hence 225% of 36 is 81

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ArbitrLikvidat [17]

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3 years ago

The sum of the angles of a triangle is 180°. If one angle of a triangle measures x and the second angle measures

serious [3.7K]

Answer:

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Step-by-step explanation:

if we subtract the two other angles the result will be the measure of the last one

4 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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3 years ago

Will mark brainliest

Lilit [14]

(2x+1)+(3x-5)=5x-4 you just have to try some of them out and see which ones work

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3 years ago