All categories

3 years ago

Can get some help on this question please?

Mathematics

2 answers:

Sergio [31]3 years ago

8 0

Answer:

(a) 8

(b) 16

(d) The ratios are the same

Step-by-step explanation:

(a) For inputs -5 and -4, the difference of outputs is ...

-3 -(-11) = 8

(b) For inputs 0 and 2, the difference of outputs is ...

45 -29 = 16

(c) For inputs -3 and 0, the difference of outputs is ...

29 -5 = 24

(d) The ratios of output difference over input difference are ...

8/1 = 16/2 = 24/3 = 8 . . . . . they are all the same

This is yet another confirmation that the slope of a line is the same everywhere, and over any interval.

zloy xaker [14]3 years ago

5 0

Answer:

Step-by-step explanation:

The input is x on the first row

The output is f(x) on the second row

a) to determine the difference of outputs of any two inputs that are 1 unit apart, let us take x = -4 and x = -5

The outputs at these inputs are -3 and - 11 respectively

Difference in output = - 3 - - 11 = 8

b) to determine the difference of outputs of any two inputs that are 2 units apart, let us take x = -3 and x = -5

The outputs at these inputs are 5 and - 11 respectively

Difference in output = 5 - - 11 = 16

c)to determine the difference of outputs of any two inputs that are 3 units apart, let us take x = -2 and x = -5

The outputs at these inputs are 13 and - 11 respectively

Difference in output = 13 - - 11 = 24

d) as the difference in interval of the input increased by 1, the difference in outputs of the corresponding inputs increased by 8. A constant increment in the interval between inputs led to a constant increment in the intervals of the corresponding outputs

You might be interested in

Please helppp!! Find the missing side . Round to the nearest tenth

olga nikolaevna [1]

Answer:

7. 7.8

Step-by-step explanation:

We can use Tan of this angle to find the missing side (in this case, the side adjacent to the angle measuring 57°)

Tan A = (opposite side)/(adjacent side)

Tan 57° = 12/x

solve for x

x(Tan 57°) = 12

x = 12/(Tan 57°)

x = 7.8 (rounded to the nearest tenth)

6 0

2 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$