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NikAS [45]
1 year ago
11

Trisha is going to make a border of icing around a circular cake with a diameter of 8 inches.

Mathematics
2 answers:
Dvinal [7]1 year ago
8 0

Answer:

Step-by-step explanation:

la b

DanielleElmas [232]1 year ago
7 0

Answer:

B.)25.12

Step-by-step explanation:

We are simply finding the perimeter of the cake

.Perimeter: π × diameter

Perimeter: 3.14 × 8

Perimeter: 25.12 inches

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Determine which statement uses an invalid argument.
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A. because in A, it flips the two conditions, which can changes the accuracy of the statement.
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3 years ago
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What is m∠D? Enter your answer in the box.
Alex73 [517]

Answer:

D = 32

Step-by-step explanation:

A+ B + E = 180 since they form a triangle

14+ 45 + E = 180

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3 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

7 0
2 years ago
A manager must select 5 delivery locations from 9 that are available. Find the number of
Dmitry_Shevchenko [17]
C^9_5=\dfrac{9!}{5!4!}=\dfrac{6\cdot7\cdot8\cdot9}{2\cdot3\cdot4}=2\cdot7\cdot9=126
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3 years ago
-3p+6=-12<br> How to solve
vfiekz [6]
Subtract 6 which leaves u with -3p=-18 then divide by three on both sides to get p=6
6 0
3 years ago
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