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tino4ka555 [31]

2 years ago

10

Hugo plans to buy equal numbers of cupcakes and cookies . a package of cupcakes contains 10 cupcakes, and a package of cookies c

antains 12 cookies

1 answer:

vova2212 [387]2 years ago

5 0

What are you asking how many in he has 22

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The figure shows the graph of f(x) as a solid curve and the graph of g(x) as a dashed curve which equation describes the relatio

Olenka [21]

Answer:

A. g(x) = f(x - 6) + 2

Step-by-step explanation:

Vertex moves from:

(-6,-2) to (0,0)

g(x) = f(x - 6) + 2

4 0

2 years ago

The record for the most rolls of wrapping paper sold for the school fundraiser is 593. Let w be the number of rolls of wrapping

astra-53 [7]

Answer:

w ≤ 593

Step-by-step explanation:

Missing option;

w ≥ 593

w > 593

w ≤ 593

593 < w

Explanation:

w ≥ 593 No

Number of wrapping paper sold never be more than 593.

w > 593 no

Number of wrapping paper sold never be more than 593.

w ≤ 593 yes

Number of wrapping paper sold will be equal or less than 593.

5 0

2 years ago

Which function models the area of a rectangle with side lengths of 2x - 4 units and x + 1 units?

aalyn [17]

Answer:

Area of rectangle, $f(x)=2x^2-2x-4$ .

Step-by-step explanation:

We are given with side lengths of a rectangle are (2x-4) units and (x+1) units. It is required to find the area of rectangle.

The area of a rectangle is equal to the product of its length and breadth. It is given by :

$A=L\times B$

Let us consider, L = (2x-4) units and B = (x+1) units

Plugging the side lengths in above formula:

$A=(2x-4)\times (x+1)$

$A=2x^2+2x-4x-4\\\\A=2x^2-2x-4$

So, the function that models the area of a rectangle is $f(x)=2x^2-2x-4$ .

5 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

2 years ago

astraxan [27]

The mode is 2.4 idk the rest

4 0

2 years ago