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yawa3891 [41]
3 years ago
6

Find the circumference and area of a circle with a diameter of 22 inches. Leave your answers in terms of pi. (4 points) C = 11π;

A = 44π C = 22π; A = 44π C = 11π; A = 121π C = 22π; A = 121π
Mathematics
2 answers:
dolphi86 [110]3 years ago
8 0
The answers are A= 121 pi D= 22 pi
alexandr402 [8]3 years ago
5 0
11×11×π=121π
A=121π
2×11×π=22π
D=22π
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Write the following as a fraction in simple form <br><br>1.6​
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It says write an equation of the line shown and the line goes through points 4,3 and 1,-3
RUDIKE [14]
Slope of the line;
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3 years ago
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]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

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]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

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Therefore the value of y(1)= 0.9152.

4 0
3 years ago
Nissan Appliances bought two dozen camcorders at a total cost of $3,816. The markup on the camcorders is 25% of the selling pric
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How to do graphs of quadratic functions ​
Anastaziya [24]

Answer:

The Graph of a Quadratic Function

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Step-by-step explanation:

Example: Graph: f(x)=−x2−2x+3.

Solution:

Step 1: Determine the y-intercept. To do this, set x=0 and find f(0).

f(x)f(0)===−x2−2x+3−(0)2−2(0)+33

The y-intercept is (0,3).

Step 2: Determine the x-intercepts if any. To do this, set f(x)=0 and solve for x.

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Here where f(x)=0, we obtain two solutions. Hence, there are two x-intercepts, (−3,0) and (1,0).

Step 3: Determine the vertex. One way to do this is to first use x=−b2a to find the x-value of the vertex and then substitute this value in the function to find the corresponding y-value. In this example, a=−1 and b=−2.

x====−b2a−(−2)2(−1)2−2−1

Substitute −1 into the original function to find the corresponding y-value.

f(x)f(−1)====−x2−2x+3−(−1)2−2(−1)+3−1+2+34

The vertex is (−1,4).

Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose x=−2 and find the corresponding y-value.

x −2  y 3f(−2)=−(−2)2−2(−2)+3=−4+4+3=3Point(−2,3)

Our fifth point is (−2,3).

Step 5: Plot the points and sketch the graph. To recap, the points that we have found are

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Answer:

4 0
3 years ago
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