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GaryK [48]
3 years ago
12

HOW MANY TRIANGLES ARE THERE? INCLDNG OVERLAPPING TRIANGLES. I NEED ANSWERS. PLEASE HELP. TYIA​

Mathematics
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

The answer is 48.

Step-by-step explanation:

There are 4 triangles in each square. So multiply 4 by 12 because thats how many squares there are.

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(x⁴)³<br>ano po answer<br>plss​
yanalaym [24]

Answer:

x^12

Step-by-step explanation:

multiply 4 and 3

x^4*3 = x^12

4 0
3 years ago
Given that a rectangle has a length of 5/2x + 10 with a width of 5/2x + 5, formulate an expression to represents the area of the
stepan [7]

Answer:

<h3>               A =  ²⁵/₄x² + ⁷⁵/₂x + 50</h3>

Step-by-step explanation:

L = ⁵/₂x + 10

W = ⁵/₂x + 5

A = L•W

A = (⁵/₂x + 10)(⁵/₂x + 5)

A = ⁵/₂x•⁵/₂x + ⁵/₂x•5 + 10•⁵/₂x + 10•5

A = ²⁵/₄x² + ²⁵/₂x + ⁵⁰/₂x + 50

A =  ²⁵/₄x² + ⁷⁵/₂x + 50

Or if yoy mean:

L = 5/(2x) + 10

W = 5/(2x) + 5

A = [5/(2x) + 10][5/(2x) + 5] = 25/(4x²) + 75/(2x) + 50

8 0
3 years ago
Please Help!!!!!!!!!
marissa [1.9K]
124 square units
you just have to count the top, bottom, and a side and times by 2
4 0
3 years ago
The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of
defon
The abscissa of the ordered pair, that is the x-coordinate, is equal to 1 and the ordinate, the y-coordinate, is equal to -1. In the cartesian plane, this point lies in the fourth (IV) quadrant. The standard position of the angle is that which has one of its side is in the x-axis.

Solve for the hypotenuse of the right triangle formed.
           h = sqrt((-1)² + (1)²) = √2
Below items show the calculation for each of the trigonometric functions.

 sin θ = opposite/hypotenuse = y/h = (-1)/(√2) = -√2/2
 cos  θ = adjacent/hypotenuse = x/h = (1)/√2 = √2/2
tan θ = opposite/adjacent = y/x = -1/1 = -1
7 0
3 years ago
A. Plot the data for the functions f(x) and g(x) on a grid and connect the points.
labwork [276]

Answer:

  a) see the plots below

  b) f(x) is exponential; g(x) is linear (see below for explanation)

  c) the function values are never equal

Step-by-step explanation:

a) a graph of the two function values is attached

__

b) Adjacent values of f(x) have a common ratio of 3, so f(x) is exponential (with a base of 3). Adjacent values of g(x) have a common difference of 2, so g(x) is linear (with a slope of 2).

__

c) At x ≥ 1, the slope of f(x) is greater than the slope of g(x), and the value of f(x) is greater than the value of g(x), so the curves can never cross for x > 1. Similarly, for x ≤ 0, the slope of f(x) is less than the slope of g(x). Once again, f(0) is greater than g(0), so the curves can never cross.

In the region between x=0 and x=1, f(x) remains greater than g(x). The smallest difference is about 0.73, near x = 0.545, where the slopes of the two functions are equal.

4 0
3 years ago
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