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2 years ago

Please find the length of the diagonal, thanks. Will give brainliest

Mathematics

2 answers:

rodikova [14]2 years ago

7 0

Answer is 12.73
Use Pythagorean theory
Split square into two right triangles with diagonal line. To find the diagonal
a^2 + b^2 = c^2
9^2 + 9^2 = c^2
162= c^2
Take square root of both sides
12.73 = c

Alborosie2 years ago

4 0

The correct answer is 12.73

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7. A city lot is twice as long as it is wide, If its perimeter is 480 units, what are the dimensions of the lot?

g100num [7]

Answer:

Step-by-step explanation:

P = 480

P = 2 length + 2 width

but we also know that

2width = length

so plug that into the length

P = 2(2width) + 2 width

480 = 4 width + 2 width

480 = 6 width

80 = width

160= length

see?

3 0

3 years ago

Andy has a collection of movie DVD's. In Andy's collection 3/5 of the DVD's are action and 1/4 of the DVD's are comedy . Andy sa

valkas [14]

Andy is wrong.
3/5 = 12/20
1/4 = 5/20
Add them up
12/20 + 5/20 = 17/20
17/20 is not equal to 4/9

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3 years ago

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2 years ago

What is the percent change from 20 to 18?

irinina [24]

<span>there was a 10% decrease</span>

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3 years ago

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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. =)

3 0

3 years ago