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3 years ago

How can you use geometric formulas in real world situations

Mathematics

1 answer:

aev [14]3 years ago

6 0

By understanding geometric symmetry and understanding natural phenomena

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Simplify (2x^2y)^3 (-3xy^2)^2

solniwko [45]

Simplified it's 576x^8y^7

8 0

3 years ago

The base of a trinagle is increased by 20% and it’s height is increased by 30%. By what percent is the area of the triangle incr

Molodets [167]

Lets say the base is x and the height is y. The new base would be 1.2x and the new height would be 1.3y

Area of a triangle is base times height divided by 2:

1.2x*1.3y/2 = (1.2*1.3 + x*y)/2 = 1.56xy/2 = 0.78xy = 78% of xy

But a normal area would be 50% of xy

78% - 50% = 38%

ANSWER: 38% increase

4 0

3 years ago

The vertices of a rectangle are R(-5, -5), S(-1, -5), T(-1, 1), and U(-5, 1). a translation, R to the point (-4,2). Find the tra

Nikitich [7]

Translation:
R ( - 5, - 5 ) → R` ( - 4, 2 )
- 4 = - 5 + 1, 2 = - 5 + 7;
The translation rule:
( x , y ) → ( x + 1, y + 7 )
Coordinates of the point U are (- 5, 1 )
- 5 + 1 = - 4, 1 + 7 = 8
The image of U is :
U` ( - 4, 8 )

4 0

3 years ago

Add.<br> (8y + 6) + (4y + 5)<br> Submit

Inga [223]

Answer:

12y + 11

Step-by-step explanation:

To add them you would add like terms. That means you would add those with y together and those without together.

8y + 4y + 6 + 5 = 12y + 11

If you want to solve for y, subtract 11 from both sides then divide both sides by 12.

6 0

3 years ago

Read 2 more answers

Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?

Agata [3.3K]

Answer:

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

6 0

2 years ago