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ch4aika [34]
3 years ago
13

B. In how many games did Katy play catcher? Is this right

Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

yea you did it right good job ;]

Step-by-step explanation:

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A rectangle had a length of 2x-7 and a with of 3x+8y. Write and simplify an expression for the perimeter of that rectangle?
wel

Answer:

Step-by-step explanation:

A rectangle had a length of 2x-7 and a width of 3x+8y

Perimeter P = 2(l + w)

P = 2(2x - 7 + 3x + 8y)

P = 2(5x + 8y - 7)

P= 10x + 16y - 14

4 0
4 years ago
What is the degree of the polynomials w3-2w+w-5
yuradex [85]
16w because you need to add it up
7 0
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Factor -8x^3-2x^2-12x-3
inysia [295]

-8x^3-2x^2-12x-3\\\\=-2x^2(4x+1)-3(4x+1)\\\\=(4x+1)(-2x^2-3)\\\\=\boxed{-(4x+1)(2x^2+3)}

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3 years ago
Eric has 7 red shirts and 11 black shirts what is the ratio of red and black shirts.
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Answer:

7:11

Step-by-step explanation:

Hope this helps! God bless.

7 0
3 years ago
Read 2 more answers
What is an equation for a sine curve with amplitude 2, and period 4pi radians ?
Nezavi [6.7K]
<span>1. a sine curve with amplitude 2, and period 4pi radians
</span>
the general equation of the sine curve ⇒⇒ y = a sin (nθ)
where: a is the amplitude  and  n = 2π/perid
∵ <span>amplitude 2, and period 4pi radians
</span>
∴ y = 2 sin (θ/2)

The correct answer is option D. y = 2 sin (θ/2)
===========================================

<span>2.The period and amplitude of the function ⇒⇒ y = 5 cos 2θ
</span>
<span>comparing with y = a cos nθ
</span>
where : a is the amplitude  and  n = 2π/period
<span>amplitude = 5  , period = 2π/n = 2π/2 = π
</span>

The correct answer is option B. Period: pi radians: Amplitude:5

============================================================
3. tan (2π/3) = tan 120° = -√3 
120° lie in the second quadrant and its reference angle = 180° - 120° = 60°
tan function in the second quadrant is negative
∴ tan 120° = - tan 60 = -√3

The correct answer is C. -sqrt3

=====================================================
4. <span>Tan 5π/6 = tan 150° = -(√3)/3
</span>
150° lies in the second quadrant and its reference angle = 180° - 150° = 30°
tan function in the second quadrant is negative
∴ tan 150° = - tan 30 = -(√3)/3

The correct answer is <span>B.-sqrt3/3</span>
8 0
3 years ago
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