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kifflom [539]
3 years ago
5

What's the answer???

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0
A was the answer I got because she checked her answer and came up with a soution.


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ABCD is a parallelogram and this figure is NOT drawn to scale.
dimulka [17.4K]
The angle does not matter. Think of it as finding the other side to a triangle. Use a ^{2} +b^2 =c^2

a=39 (line AB)
b=b (the leg we need to find)
c=89 (line BD)

39^2 + b^2 = 89^2
1521 + b^2 = 7921
(subtract the 1521 from both sides)
b^2 = 6400
(square root both sides)
\sqrt{(b^2)} =  \sqrt{6400} 


b = 80
AD=80
7 0
3 years ago
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The graph of f(x) = x2 + x - 6 is shown below.
guapka [62]

Answer:

the solutions of a function are the points where for some value of x the function becomes zero

thus the solns for this graph would be

<h3>-3 , 2</h3>

that's option 1.

3 0
2 years ago
Explain why the polynomial 100x^2 + 150x + 49 is not a perfect square.
Ad libitum [116K]
Answer: The correct answer is Choice C.

For this polynomial to be a perfect square, it would need to be:
(10x + 7)^2

This will ensure that the first terms and the last terms will be 100x^ and 49. However, if you use foil to multiply the factors, you will not get 150x for the center term. Choice C also states that 150x will not be the middle term.
4 0
3 years ago
Find the radius x of OC.<br> a) 4 <br> b) 6<br> c) 12<br> d) 12.8
Dahasolnce [82]

Answer:

the correct answer is 6..... option b

5 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
3 years ago
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