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IRISSAK [1]
3 years ago
12

A red light flashes every 14 minutes. A blue light flashes every 24 minutes. when will the two lights flash together again if th

ey last flashed together at 8 AM?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0
10:48 am (168 minutes later)
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What’s the area of a circle with a diameter of 4
umka21 [38]

Answer:

4\pi ≈ 12.57 units squared

Step-by-step explanation:

The area of a circle is denoted by: A=\pi r^{2}, where r is the radius.

Remember that diameter is twice the radius, so since the diameter equals 4, we know that the radius r = 4/2 = 2.

Now, plug this in: A=\pi *2^2=4\pi

We can round 4pi to 12.57 units squared.

Hope this helps!

7 0
3 years ago
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SIZIF [17.4K]
38. Is the answer. Go go go!
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3 years ago
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18-(∛27)^3<br> Find The cube root
pickupchik [31]
The cube root is - 9
7 0
2 years ago
Which inequality hs the graph shown below. (will mark you brainliest)
zvonat [6]

Answer:

y < 2x-3

Step-by-step explanation:

First you graph the -3, and the slope is 2 so you go up 2 and over 1. Since y is less than 2x-3, you shade the right region.

6 0
2 years ago
Perform a first derivative test on the function ​f(x)equals2 x cubed plus 3 x squared minus 120 x plus 6​; ​[minus5​,8​]. Bold
maria [59]

Answer:

a) Critical points

x = 4 and x = -5

b) x = 4 corresponds to a minimum point for the function f(x)

x = - 5 corresponds to a maximum point for the function f(x)

c) The minimum value of f(x) in the interval = -298

The maximum value of f(x) in the interval = 431;

Step-by-step explanation:

f(x) = 2x³ + 3x² - 120x + 6 in the interval [-5, 8]

a) To obtain the critical points, we need to obtain the first derivative of the function with respect to x. Because at critical points of a function, f(x), (df/dx) = 0

f'(x) = (df/dx) = 6x² + 6x - 120 = 0

6x² + 6x - 120 = 0

Solving the quadratic equation,

x = 4 or x = -5

The two critical points, x = 4 and x = -5 are in the interval given [-5, 8] (the interval includes -5 and 8, because it's a closed interval)

b) To investigate the nature of the critical points, we obtain f"(x)

Because f'(x) changes sign at the critical points

If f"(x) > 0, then it's a minimum point and if f"(x) < 0 at c, then it's a maximum point.

f"(x) = (d²f/dx²) = 12x + 6

at critical point x = 4

f"(x) = 12x + 6 = 12(4) + 6 = 54 > 0, hence, x = 4 corresponds to a minimum point.

at critical point x = -5

f"(x) = 12x + 4 = 12(-5) + 4 = -56 < 0, hence, x = -5 corresponds to a maximum point.

c) At x = 4,

f(x) = 2x³ + 3x² - 120x + 6 = 2(4)³ + 3(4)² - 120(4) + 6 = -298

At x = - 5

f(x) = 2x³ + 3x² - 120x + 6 = 2(-5)³ + 3(-5)² - 120(-5) + 6 = 431

7 0
2 years ago
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