Which statement is true?

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

5 0

Answer:

Step-by-step explanation:

bonufazy [111]3 years ago

4 0

Answer: D

Step-by-step explanation:

A function is a relation where each input value is assigned to only one output value.

The domain of a function is the set of all input values, or x-values, for which the function is defined.

The range of a function is the set of all output values, or y-values, for which the function is defined.

Which A, B, & C are wrong. So D, is the one that makes the most sense.

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Please help me with these problems; geometry is really getting on my nerves. T-T

ivolga24 [154]

For #19 part A :

By vertical angles and then same side interior angles 3x + 2x = 180
5x = 180
x = 36

For #19 part B:

You are correct with the 35, but you missed the fact that the 105 angle is supplementary with the unknown angle inside the triangle. So:

180 - 105 = 75

Then:

180 = 35 + 75 + x
180 = 110 + x
x = 70

Last question:

Do you know what the symbol means? Text me and tell me and I can answer it for you. :)

7 0

3 years ago

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A triangular plaque has side lengths of 8 inches, 13 inches, and 15 inches.

lianna [129]

<h3>Answer: A) 52 square inches</h3>

==========================================================

Explanation:

a = 8, b = 13, and c = 15 are the sides of the triangle

s = (a+b+c)/2 = (8+13+15)/2 = 18 is the semi-perimeter, aka half the perimeter.

Those values are then plugged into Heron's Formula below

$A = \sqrt{s*(s-a)*(s-b)*(s-c)}\\\\A = \sqrt{18*(18-8)*(18-13)*(18-15)}\\\\A = \sqrt{18*(10)*(5)*(3)}\\\\A = \sqrt{2700}\\\\A = \sqrt{900*3}\\\\A = \sqrt{900}*\sqrt{3}\\\\A = 30\sqrt{3}\\\\A \approx 51.9615\\\\A \approx 52\\\\$

The triangular plaque has an area of approximately 52 square inches.

6 0

2 years ago

Calculate the average travel time for each distance, and then use the results to calculate. A 6-column table with 3 rows in the

BigorU [14]

The average travel time is simply the mean travel time between the given points

The average time that it takes for the car to travel the first 0.25m is 2.23s.
The average time to travel just between 0.25m and 0.50m is 0.9s.
Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.

<u>(a) The average time to travel the first 0.25m</u>

The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.

So, the average time to travel is:

$Time = \frac{2.24s + 2.21s + 2.23s}{3}$