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Advocard [28]
3 years ago
9

What is the first step in solving the quadratic equation -5x + 8 = 1337

Mathematics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

Jus took the test and the answers D

Step-by-step explanation:

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N is an integer.<br>Write the values of n such that -15 &lt; 3 &lt; 6​
Thepotemich [5.8K]

Answer:

Step-by-step explanation:

n could be : -14, -13, -12... -1, 0 , 1, 2, 3 ,4,5

3 0
3 years ago
15 pts! Will Give Brainliest Halp! PLZ
aivan3 [116]

Theres a saying: Dividing fractions dont ask why, just flip the second and multiply.

so a/b / c/d = a/b * d/c

-8/2 * -3/6

1. cross - cancel common factor 2

4/2 * -3/3

2. Multiply

-4(-3)/ 2*3

3. multiply numbers

--12/2*3

- -12/6

4. Apply fraction rule -a/b=- a/b

=-(-12/6)

5. divide

= -(-2)

6. Apply rule -(-a)=a

=2

Your answer would be 2


3 0
4 years ago
Read 2 more answers
How do u do this step by step ? -8 ÷ 3.2​
Ad libitum [116K]

Answer:

the answer is -2.5

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Consider the line 3x - 4 y = 7.
Trava [24]

Answer:

The slope of a line perpendicular to this line is -1/3. The slope of a line parallel to this line would be 3.

Step-by-step explanation:

Perpendicular lines have opposite(sign) and reciprocal slopes.

Parallel lines have the same slope.

3 0
1 year ago
Let f be the function defined by f(x) = e^(x) cos x.
Pavel [41]
(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

   \displaystyle&#10;f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}

____________

(b)

f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\&#10;f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\&#10;f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}

The slope of the tangent line is e^{3\pi/2}.

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\&#10;0 = e^x \big( \cos(x) - \sin(x)\big)

Use zero factor property to solve.

e^x \ \textgreater \  0\forall x \in \mathbb{R} so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

\cos (x) - \sin (x) = 0 \\&#10;\cos(x) = \sin(x)

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\&#10;x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]

We check the values of f at the end points and these two critical numbers.

f(0) = e^1 \cos(0) = 1

\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4}  \frac{\sqrt{2}}{2}

\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4}  \frac{-\sqrt{2}}{2} = -e^{\pi/4}  \frac{\sqrt{2}}{2}

f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}

There is only one negative number.
The absolute minimum value of f <span>on the interval 0 ≤ x ≤ 2π is
-e^{5\pi/4} \sqrt{2}/2

____________

(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, \lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0

g is a differentiable function; therefore, it is a continuous function, which tells us \lim_{x \to \pi/2} g(x) = g(\pi/2) = 0.

When we observe the limit  \displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}, the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}

f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\&#10;g'(\pi/2) = 2

thus

\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}</span>

3 0
3 years ago
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