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Luba_88 [7]
3 years ago
4

Rectangle ABCD with coordinates A(1, 1), B(4, 1), C(4, 2), and D(1, 2) dilates with respect to the origin to give rectangle A'B'

C'D'. If A'B' = 6, what is the scale factor of the dilation?
a.6
b.3
c.2
d.4
Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0
I would choose a for my option. Hope I helped thxs
Firdavs [7]3 years ago
4 0
The answer is d...........



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Joe barrowed $2,000 from the bank at a rate of 7% simple interest per year. How much will be in your account in 5 years?
timofeeve [1]
For the answer to the question above,
The simple interest formula is I = P * r * t

P is the principal (2,000)
r is the interest rate (7% or 0.07)
t is the time (5)

2000 * 0.07 * 5 = 700

$700.00 is the interest

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I hope this helps
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A city planner is rerouting traffic in order to work on a stretch of road. The equation of the path of the old route can be desc
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Answer:

<em>y - P = -5/2 (x - Q)</em>

Step-by-step explanation:

on flvs geometry module 4 test, this is correct!

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3 years ago
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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

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Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

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\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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Step-by-step explanation:

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