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3 years ago

What is the range of f(x)=43*2.4^x

Mathematics

1 answer:

Aliun [14]3 years ago

7 0

This is an exponential function.
Without any transformations (up or down), the range is y > 0

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PLEASE HELP !!!!!!!!!!!!

V125BC [204]

The answer would be b - associate w it

3 0

3 years ago

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago

What is the slope of the line shown in the graph? (4 points)

Korvikt [17]

Answer:

a. or $\frac{3}{-2}$

Step-by-step explanation:

Slope is equal to y1-y2 over x1-x2.

So the equation would be:

$\frac{2-(-1)}{-3-(-1)}$

2-(-1)=2+1=3

-3-(-1)=-3+1=-2

So the answer is:

$\frac{3}{-2}$

5 0

3 years ago

-1/2(2x-6)+9=-8 answer

Oxana [17]

Answer:

x= 20 i think sorry if it's wrong⊙﹏⊙

5 0

3 years ago

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40 divided by(-0.25)

Allushta [10]

Answer:

-160

Step-by-step explanation:

6 0

3 years ago

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