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Inessa05 [86]
3 years ago
8

Which of the following are measurements for triangles that are similar to a triangle with sides measuring 6, 8, and 12? Check al

l that apply
A. 3, 4, and 6

B. 18, 24, and 36

C. 2, 3, and 4

D. 4.8, 6.4, 9.6

E. 14.4, 20.8, and 36

you can choose more than one answer
Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
6 0

Answer:

A, B, D

Step-by-step explanation:

Use the LCM to help...

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Since, Frank needs to save $600 to buy a set of golf clubs. He plans to save $75 per month.

Now, we have to determine the amount of money still he has to save (y) in relation to the number of months (x) in which he has saved money.

Let 'y' be the amount of money he still have to svae.

Let 'x' be the number of months he has saved money.

Total money saved yet = $75 \times x= $75x

He has to save $600 in total.

So, Money he still have to save = 600 - 75x

So, y=600-75x is the required equation.

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3 years ago
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A home improvement store advertises 60 square feet of flooring for $287.00 including an $80.00 installation fee. How much does e
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$287-80 for installation leaves $207 for flooring cost
207/60 = $3.44/square foot
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3 years ago
Enola owns a small business selling ice-cream. She knows
S_A_V [24]

Answer:

5.65................................

Step-by-step explanation:

6 0
2 years ago
Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =
Helen [10]

For the ODE

ty'+2y=\sin t

multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:

t^2y'+2ty=t\sin t

\implies(t^2y)'=t\sin t

Integrate both sides with respect to <em>t</em> :

t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C

Divide both sides by t^2 to solve for <em>y</em> :

y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}

Now use the initial condition to solve for <em>C</em> :

y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}

\implies9=\dfrac4{\pi^2}(1+C)

\implies C=\dfrac{9\pi^2}4-1

So the particular solution to the IVP is

y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}

or

y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}

6 0
3 years ago
Consider w=sqrrt2/2(cos(225°) + isin(225°)) and z = 1(cos(60°) + isin(60°)). What is w+ z expressed in rectangular form?
SVETLANKA909090 [29]

Answer:

Option (3)

Step-by-step explanation:

w = \frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]

Since, cos(225) = cos(180 + 45)

                          = -cos(45) [Since, cos(180 + θ) = -cosθ]

                          = -\frac{\sqrt{2}}{2}

sin(225) = sin(180 + 45)

             = -sin(45)

             = -\frac{\sqrt{2}}{2}

Therefore, w = \frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]

                      = -\frac{2}{4}(1+i)

                      = -\frac{1}{2}(1+i)

z = 1[cos(60) + i(sin(60)]

  = [\frac{1}{2}+i(\frac{\sqrt{3}}{2})

  = \frac{1}{2}(1+i\sqrt{3})

Now (w + z) = -\frac{1}{2}(1+i)+\frac{1}{2}(1+i\sqrt{3})

                   = -\frac{1}{2}-\frac{i}{2}+\frac{1}{2}+i\frac{\sqrt{3}}{2}

                   = \frac{(i\sqrt{3}-i)}{2}

                   = \frac{(\sqrt{3}-1)i}{2}

Therefore, Option (3) will be the correct option.

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3 years ago
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