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Mandarinka
13 days ago
11

Lin created a scaled copy of Triangle A with an area of 72 square units. How many times larger is the area of the scaled copy co

mpared to that of Triangle A

Mathematics
2 answers:
PIT_PIT [3.9K]13 days ago
6 0

Answer:

The question appears to be incomplete, here’s a possible interpretation of the full question:

Here is Triangle A. Lin produced a scaled version of Triangle A with an area of 72 square units. What scale factor was used by Lin to create this copy? Recall: A=1/2bh

a) 4

b) 8

c) 16

Answer:

Scale factor = 16

Step-by-step explanation:

Based on the diagram provided, the triangle was drawn on graph paper, where each grid equals 1 unit. Thus, the dimensions of Triangle A derived from the diagram are:

Base = 3 units

Height = 3 units

To find the scale factor associated with the area of the triangle after scaling, we first calculate the area of the original triangle.

Area of Triangle = 1/2 (base × height)

Area of Triangle = 0.5 × 3 × 3 = 4.5 square units

Consequently,

Area of the original triangle = 4.5 square units

Area of the scaled triangle = 72 square units

Since the area of the scaled triangle exceeds that of the original, the scale factor simply reflects how many times the area was increased from the original triangle to the scaled one, calculated as:

Scale factor = (scaled triangle) ÷ (original triangle)

Scale factor = 72 ÷ 4.5 = 16

Leona [4.1K]13 days ago
0 0

What is the proportion by which the area of the scaled version is larger compared to Triangle A?

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Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 co
PIT_PIT [3949]

Answer:

u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt.

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

Thus, Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2).

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Initial conditions yield:

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

Which leads us to C1 + 8 /(1- w^2) = 5

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Next, taking the derivative:

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

   

4 0
2 days ago
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Henry needs 2 pints of red paint and 3 pints of yellow paint to get a specific shade of orange. If he uses 9 pints of yellow pai
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Answer:

In total, Henry requires 15 pints of paint.

Step-by-step explanation:

Henry starts with 2 pints of red and 3 pints of yellow to achieve a certain color, but he has already used 9 pints of yellow paint. Thus, this is represented by:

3 x 3 = 9 (yellow)

Since whatever is done on one side must also be applied to the other, we calculate the red paint:

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Now we consolidate the amounts:

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Here, 'a' relates to 0.

There are two scenarios for 'r' and 't'.

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Both are positioned on the same side to the right of 'a'.

In this case, 'r' would equal 5, and 't' would equal 7.

The midpoint between 'r' and 't' is \frac{5+7}{2} =6.

Scenario 2.

If both are found to the left of 'a'.

Then 'r' would equal -5, while 't' would equal -7.

The midpoint is \frac{-7-5}{2} =-6.

Scenario 3.

If 'r' is right of 'a' and 't' is left of 'a'.

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Scenario 4.

If 'r' is left of 'a' while 't' is right of 'a'.

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The midpoint is \frac{-5+7}{2}=1.

The potential midpoint coordinates for 'rt' are 6, -6, 1, and -1.

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