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aniked [119]
2 years 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:
Maurinko [17]2 years ago
6 0

Answer:

The question is not complete, here is a possible match to the complete question:

Here is Triangle A. Lin created a scaled copy of Triangle A with an area of 72 square units. What scale factor did Lin apply to the Triangle A to create the copy? Remember: A=1/2bh

a) 4

b) 8

c) 16

Answer:

Scale factor = 16

Step-by-step explanation:

From the diagram attached to this solution, the triangle was plotted on a graph sheet, and each grid on the graph represents 1 unit. hence the dimensions of Triangle A from the diagram is as follows:

Base = 3 units

Height = 3 units

Next, in order to determine the scale factor of the area of the triangle after scaling, let us calculate the area of the unscaled triangle.

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

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

Therefore,

Area of unscaled triangle = 4.5 squared units

Area of scaled triangle = 72 squared units

since the area of the scaled triangle is larger than the unscaled triangle, the scale factor is simply the number of times by which the scaled triangle was enlarged, compared to the unscaled triangle. This can be calculated by dividing the scaled triangle by the unscaled triangle as follows:

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

Scale factor =  72 ÷ 4.5 = 16

Guest1 year ago
0 0

How many times larger is the area of of the scaled copy compared to that of Triangle A?

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VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

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Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

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the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

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y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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1 year ago
A volunteer is organizing the snacks at a concession stand and notices the ratio is sweet snacks to salty snacks is 5:1. If ther
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Answer:

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In other words, to compute the next term in the series you have to multiply the previous one by r.

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots.

Let's use the other information, since the last term is 4374 > 6, we know that r>1, otherwise the terms would be bigger and bigger.

The information about the sum tells us that

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We have a formula to compute the sum of the powers of a certain variable, namely

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