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slavikrds [6]
3 years ago
6

Express each ratio as a unit rate.round to the bearest tenth,if necessary. 2 note books for $0.60

Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0

Answer:

(B) $0.30 per notebook

Step-by-step explanation:

FrozenT [24]3 years ago
6 0

Option B

2 notebooks for $ 0.60 expressed as unit rate is $ 0.30 per notebook

<h3><u>Solution:</u></h3>

We have to express ratio as unit rate

Given that, 2 notebooks for $ 0.60

Number of notebooks = 2

Cost of 2 notebooks = $ 0.60

So we have to find cost of 1 notebook

\begin{array}{l}{\text { cost of } 1 \text { notebook }=\frac{\text { cost of } 2 \text { notebook }}{2}} \\\\ {\text { cost of } 1 \text { notebook }=\frac{0.60}{2}=0.30}\end{array}

So unit rate = cost of 1 notebook = $ 0.30

So 2 notebooks for $ 0.60 as unit rate is $ 0.30 per notebook

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Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea
Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

y=y_1u_1+y_2u_2

where

u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}

Then

u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t

u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}

The general solution to the ODE is

y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}

which simplifies somewhat to

\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}

4 0
3 years ago
A triangle is removed from the rectangle. Find the area of the remaining figure.
Mariulka [41]

Answer

C) 160 in^2


Step by step explanation

First, let's find the area of the rectangle.

Area of a rectangle = length x width

The area of the rectangle = 20 * 10

= 200 in^2

Now let's find the area of the triangle that is removed.

Area of a triangle = 1/2 base * height.

Here base = 10 and height = 8. Plug in these values into the formula, we get

Area of the triangle = 1/2 * 10*8

= 5* 8

The area of the triangle = 40 in^2

The area of the remaining figure = Area of the rectangle - area of the triangle

=  200 - 40

= 160 in^2

The answer is "160 in^2"

Thank you.

5 0
3 years ago
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Furkat [3]
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7 0
3 years ago
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225% of what number is 81
Airida [17]
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225% = 225 out of 100 = 225/100
So we have this equation, where we call x the number we are looking for:
<span>(225%)(x) = 81
</span>(225/100)(x) = 81
so we solve for x:
225x = 81*100 = 8100
x = 8100/225
x = 36
The original number is 36, hence 225% of 36 is 81
4 0
3 years ago
If two sides of a triangle are 5 and 7 inches longer than the third side and the perimeter measures 21, find the length of each
aleksandr82 [10.1K]
Hello

x
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3 x = 9
x = 3 

3 , 5 et 7 pouces
7 0
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