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Helga [31]
2 years ago
11

Find the linear inequality & show it on a number line: 5(x-3) < 20

Mathematics
1 answer:
IRINA_888 [86]2 years ago
6 0

Answer:

Step-by-step explanation:

5(x-3)<20

5x-15<20

5x<35

x<7

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Let f be a differentiable function on (-0o,00) such that f(-x)= f(x) for all x in (, o). Compute the value of f'(0). Justify you
snow_lady [41]

Answer:

f'(0)=0

Step-by-step explanation:

Applying the chain rule

\frac{d}{dx} (f(-x))=-\frac{df}{dx}

Then it becomes

\frac{df}{dx} =-\frac{df}{dx}

In x=0

\frac{d[tex]f'(0)=-f'(0)\\f'(0)+f'(0)=0\\2f'(0)=0\\f}{dx} =-\frac{df}{dx}[/tex]

Then

f'(0)=0

8 0
3 years ago
Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2) , and X(1, −2) . Trapezoid WKLX ​ is translated 4 units right and 3 uni
IRINA_888 [86]
<span>WKLX
W(2, −3),  
K(4, −3),
L(5, −2) ,
X(1, −2) 

TRANSLATED 4 UNITS RIGHT and 3 UNITS DOWN to produce W'K'L'X

4 units right means the x coordinate is affected. Since the moving to the right, we add 4 to the x values of each vertice.

W = 2 + 4 = 6
K = 4 + 4 = 8
L = 5 + 4 = 9
X = 1 + 4 = 5

3 units down means the y axis is affected. We add 3 to the value of y but keep the negative sign.
W = -3 + -3 = -6
K = -3 + -3 = -6
L = -2 + -3 = -5
X = -2 + -3 = -5

The correct answer is: </span><span>W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)</span>
6 0
3 years ago
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