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3 years ago

Lowest to highest 7.2 ; 2.07 ; 7.02 ; 27.2 ; 0.27

Mathematics

2 answers:

dybincka [34]3 years ago

8 0

Answer:

0.27 ; 2.07 ; 7.02 ; 7.2 ; 27.2

DaniilM [7]3 years ago

7 0

Answer:

27.2 , 7.2 , 7.02, 2.07, 0.27

Step-by-step explanation:

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If the squared difference of the zeroes of the quadratic polynomial x2+kx+30 is equal to 169 find the value of k and the zeroes

anyanavicka [17]

ANSWER

$x = 2 \: \: or \: \: x = 15$

$x = - 2 \: \: or \: \: x = - 15$

EXPLANATION

The given polynomial is

$f(x) = {x}^{2} + kx + 30$

where a=1,b=k, c=30

Let the zeroes of this polynomial be m and n.

Then the sum of roots is

$m + n = - \frac{b}{a} = -k$

and the product of roots is

$mn = \frac{c}{a} = 30$

The square difference of the zeroes is given by the expression.

$( {m - n})^{2} = {(m + n)}^{2} - 4mn$

From the question, this difference is 169.

This implies that:

$( { - k)}^{2} - 4(30) = 169$

${ k}^{2} -120= 169$

$k^{2} = 289$

$k= \pm \sqrt{289}$

$k= \pm17$

We substitute the values of k into the equation and solve for x.

$f(x) = {x}^{2} \pm17x + 30$

$f(x) = (x \pm2)(x \pm 15)$

The zeroes are given by;

$(x \pm2)(x \pm 15) = 0$

$x = \pm2 \: \: or \: \: x = \pm 15$

5 0

3 years ago

The length of a side of a square whose perimeter is r units

Bezzdna [24]

Answer: r/4 units

Step-by-step explanation: Each side of a square is of the same length, perimeter is the sum of the 4 length, length of a side of square = r/4 units

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2 years ago

45 degrees Fahrenheit to Celsius

TiliK225 [7]

7.2
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4 years ago

PLEASE HELP ME. WILL GIVE BRAINLIEST Simplify -10x - 2x + 3x. -9x 9x 11x

s344n2d4d5 [400]

Your answer is -9x I'm 100% sure

I hope this helps

7 0

3 years ago

Read 2 more answers

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

$\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

____________

(b)

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

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

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)$

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

$\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)$

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):

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

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 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.

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

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2$

thus

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

3 0

3 years ago