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Montano1993 [528]
2 years ago
11

there are nine children and Kenneth preschool class three of them chose to play at the sand table based on past data is 12 presc

hoolers are at school today how many should you expect to play at the sand table​
Mathematics
1 answer:
ankoles [38]2 years ago
3 0

Answer:

4 are playing at the sand table

Step-by-step explanation:

There were 9 kids at Kenneth's preschool and 3 were playing, thats one third of the kids(3x3=9). One third of 12 is 4(3x4=12) Basically since there were 9 total kids, and 3 at the sand table, they are divided equally in groups of three so you just divide 12 by 3 too and thats how you get 4

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If you reflect (–2, –8) across both axes, which quadrant will it be in? Justify your reasoning.
stepan [7]
The quadrant would be the first one, because once reflected over the y-axis, it would become (2,-8), and once reflected over the x-axis, it would become (2,8)
8 0
3 years ago
Read 2 more answers
Please help I’m so stressed over this ok here it is it cost $12.50 to enter an amusement park and $1.50 to play each game. Solve
jeyben [28]
Answer:
He can go on 15 gams
3 0
2 years ago
The length of a rectangular park is four timesfour times its width. The park is surrounded by a 33​-foot-wide path. Let x denote
tiny-mole [99]

Answer:

A(x) = 4x^2 + 330x + 4356

Step-by-step explanation:

width of park: x

length of park: 4x

width + path width = x + 66

length + path width = 4x + 66

area = (x + 66)(4x + 66)

area = 4x^2 + 66x + 264x + 4356

area = 4x^2 + 330x + 4356

A(x) = 4x^2 + 330x + 4356

7 0
3 years ago
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS
Elina [12.6K]

You want to end up with A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:

A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with 2\sin(4\pi t)+5\cos(4\pi t). Right away, we know \omega=4\pi.

We also need to have

\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that \sin^2x+\cos^2x=1 for all x; this means

(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}

Then

\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)

So we end up with

2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)

8 0
3 years ago
Read 2 more answers
What are the two missing sides of the triangle ??
igor_vitrenko [27]
Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.

Givens
a = x
b = x + 4
c = 20

Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2

Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20  Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0  
2x^2 + 8x - 4 = 0         Divide through by 2
x^2 + 4x - 2 = 0

Use the quadratic formula
a = 1
b = 4
c = - 2

\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} 
 
\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} + 4(1)(2) } }{2} 


From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6)  This is an extraneous root. No line can be minus.

x1 = + 0.4495
x2 = x + 4 = 4.4495
6 0
3 years ago
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