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inn [45]
4 years ago
8

!!!PLEASE HELP ILL MARK U BRAINLIEST!!!!!! LOOK AT PICTURE!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
faltersainse [42]4 years ago
7 0

Since this is a square, we just have to square one side.

Area = (y + 1)^2. Distribute the right side.

Area = y^2 + y +y + 1. Simplify.

Area = y^2 + 2y + 1

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PLZZ HELPP!!!Which statements are true about the circle shown below? Select three options.
AURORKA [14]

Answer:

If Pi is approximated to 3.14, the circumference of the circle is 25.12 cm.

Step-by-step explanation:

Just do it - Nike

4 0
3 years ago
Write the equation of the line that is perpendicular to 5x + 3y = -21 and
Marta_Voda [28]

Answer:

5x + 3y = -21

Step-by-step explanation:

i think sorry if its wrong

3 0
2 years ago
There are 70 historical fiction books in the school library. Historical fiction books make up 1/10 of the library's collection.
Gre4nikov [31]

Answer:

There are 700 books in Library Collections.

Step-by-step explanation:

Given:

Number of historical books = 70

The equation used to find the totals number of books in library collection.

\frac{1}{10}b = 70

where b⇒ Total number of books in library collections.

We need to find the Total number of books in library collections.

Solution:

\frac{1}{10}b = 70

Now by solving the above equation we get;

First we will multiply both side by 10 we get;

\frac{1}{10}b \times 10= 70\times 10\\\\b =700

Hence there are 700 books in Library Collections.

6 0
4 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
A+b-c=180 Solve the equation
VladimirAG [237]
You can do it in multiple ways but mine is....     140+120-160=180
3 0
3 years ago
Read 2 more answers
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