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3 years ago

12

I’m sort of confused on how to state and find the domain and range of this graph

1 answer:

Yuliya22 [10]3 years ago

5 0

You have a special type of function, so you have to express the domain and range of each interval. Every aspect of this graph is closed, so you will only use brackets when writing your answers. Hence, you get [30,5] for the first interval, since there is a horizontal distance of 30 and a vertical distance of 5. You repeat the same process for each interval.

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The perimeter of a rectangular field is 342 m. If the width of the field is 76 m, what is it’s length?

Klio2033 [76]

Answer:

The length is 95 m

Step-by-step explanation:

P = 2(l+w) where l is the length and w is the width

342 = 2(l+76)

Divide each side by 2

342/2 = 2/2(l+76)

171= l+76

Subtract 76 from each side

171- 76 = l+76-76

95 = l

6 0

3 years ago

1 3/10 x 18 please show work and answer

klasskru [66]

Answer:

Step-by-step explanation:

1 3/10 x 18

(1 x 10 +3)/10 = 13/10

13/10 x 18 = (13 x 18) / 10 = 234/10 = (234/2) / (10/2) = 117/5 = 23 2/5 (a fraction is a division)

7 0

3 years ago

Gemiola [76]

The derivative of $f(x) = 2\cdot x^{2}-9$ is $f'(x) = 4\cdot x$ .

In this exercise we must apply the definition of derivative, which is described below:

$f'(x) = \lim_{x \to 0} a_n \frac{f(x+h)-f(x)}{h}$ (1)

If we know that $f(x) = 2\cdot x^{2}-9$ , then the derivative of the expression is:

$f'(x) = \lim_{h \to 0} \frac{2\cdot (x+h)^{2}-9-2\cdot x^{2}+9}{h}$

$f'(x) = 2\cdot \lim_{h \to 0} \frac{x^{2}+2\cdot h\cdot x + h^{2}-2\cdot x^{2}}{h}$

$f'(x) = 2\cdot \lim_{h \to 0} 2\cdot x + h$

$f'(x) = 4\cdot x$

The derivative of $f(x) = 2\cdot x^{2}-9$ is $f'(x) = 4\cdot x$ .

We kindly invite to check this question on derivatives: brainly.com/question/23847661

4 0

3 years ago

Find the absolute maximum and minimum values of the function f(x, y) = xy2 − y 2 − x on the region d where d = {(x, y) | x ≥ 0,

Serjik [45]

Your answer is b because something

6 0

4 years ago

The position of an object at time t is given by s(t) = -9 - 3t. Find the instantaneous velocity (by using the limit) at t = 8 by

Ede4ka [16]

Answer: $v=-3$

Step-by-step explanation:

Given

The position of a particle is $s(t)=-9-3t$

Velocity is given by the derivative of position with respect to time

$\Rightarrow v=\dfrac{ds}{dt}\\\Rightarrow v=-3$

So, velocity is constant at any point in time.

At t=8, it is $v=-3$

5 0

3 years ago