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Veronika [31]
2 years ago
13

4(2y+3)-2*2. Y=3 Pls help

Mathematics
2 answers:
JulijaS [17]2 years ago
8 0

Answer:

4(2×3+3)-2×2

4(9)-4

36-4

32

jeka57 [31]2 years ago
7 0

Answer:

Hi! The correct answer is 32!

Step-by-step explanation:

Simplify the expression.

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3a. The graph shows a school's profit P for selling x lunches on one day. PART A: What is the school's profit for selling 120 lu
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Find the value of B in the following equation if you know it goes through the point (-5, -8):
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3 years ago
For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
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