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yarga [219]
2 years ago
5

Monique came home from school to find a bowl of 5 apricots and 4 plums on the table. She decides to have a snack. First she sele

cts one and then puts it back. She then selects another. What is the probability both selections were apricots
Mathematics
1 answer:
posledela2 years ago
5 0

5/(4+5)

5/9

This is the probability.

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Square of a binomial x^2+2x+1
garik1379 [7]
To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square. 
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial. 
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2. 
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial. 
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1. 
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x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.
3 0
3 years ago
Decimals that have the same value are
shepuryov [24]
<span>Decimals that have the same value are equivalent decimals. Equivalent decimals are decimal numbers that have the same value (the same amount). In other words equivalent decimal numbers have the same value but different number of decimals. For example: 
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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

——————————

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
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Can you help me solve for x
astraxan [27]
X=2 I believe but I need to keep typing to make it longer
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2 years ago
Write the​ part-to-whole relationship for the fraction given below.
vivado [14]

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