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Iteru [2.4K]
3 years ago
10

Solve the equation.

Mathematics
1 answer:
weeeeeb [17]3 years ago
6 0

Inverse Functions. An inverse function goes the other way! Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 .

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Enter the output values (Y) with the given function and input values (X).<br> y = 3x - 2
marishachu [46]

Answer:

hilghkiii

Step-by-step explanation:

5 0
3 years ago
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A rectangular prism has a length of 3 1/2 in., a width of 3 1/2 in., and a height of 7 in. Danny has a storage container for the
marta [7]
The difference is 4.25in^3
6 0
3 years ago
What number needs to be added to both sides of the equation to complete the square? X^2−7x=5 Enter your answers in the boxes.
Andrew [12]

ANSWER

We need to add half the coefficient of x all square.

\frac{49}{4}


EXPLANATION


Given the equation,

x^{2} -7x=5.

We compare to the general equation,

ax^2+bx+c=0

This means that;

a=1,b=-7,c=-5


To complete the square we add, (\frac{b}{2})^2


That is (\frac{-7}{2})^2=\frac{49}{4}.


Hence the number to be added to both sides is \frac{49}{4}.


Note that before we complete the square, the coefficient of the term inx^2 must always be 1.

7 0
3 years ago
Arliss has two pieces of carpet runner. One is2/13 yards long and the other3/13
Lunna [17]
Solution steps:

1. Add the two given fractions.

2. Subtract the sum from 10 to find your answer.
3 0
3 years ago
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Let f be the function defined by f(x) = 3x^5-5x^3+2
sveticcg [70]
(a) When f is increasing the derivative of f is positive.

    f'(x) = 15x^4 - 15x^2 > 0
            15x^2(x^2 - 1)> 0
               x^2 - 1    > 0 (The inequality doesn't flip sign since x^2 is positive)
               x^2 > 1
    Then f is increasing when x < -1 and x > 1.

(b) The f is concave upward when f''(x) > 0.
    f''(x) = 60x^3 - 30x > 0
           30x(2x^2 - 1) > 0
             x(2x^2 - 1) > 0
            x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0

    There are four regions here. We will check if f''(x) > 0.
    x < -1/sqrt(2):     f''(-1) = -30 < 0
    -1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
    0 < x < 1/sqrt(2):  f''(0.5) = -7.5 < 0
    x > 1/sqrt(2):      f''(1) = 30 > 0

    Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
    Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).

(c) The horizontal tangents of f are at the points where f'(x) = 0
    15x^2(x^2 - 1) = 0
    x^2 = 1
    x = -1 or x = 1
    f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
    f(1) = 3(1)^5 - 5(1)^3 + 2 = 0

    Therefore, the tangent lines are y = 4 and y = 0.
8 0
4 years ago
Read 2 more answers
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