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soldi70 [24.7K]
3 years ago
7

C. Sophia Thompson is 12 years old. Her

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

43

Step-by-step explanation:

the equation would be:

age = 4x -5 (x being Sophie's age)

then you plug in the 12, and PEMDAS requires that you do the multiplication first, so it would become:

age = 48 - 5

then, you subtract to get:

age = 43

Vsevolod [243]3 years ago
3 0

Answer:

28 I think

Step-by-step explanation:

12-5=7

7x4=28

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Need your help now please; the assignment is due tonight and this is the last problem I am having trouble with. The red box is t
Kobotan [32]

The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

here:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}

y = 4x

Now, our region bounded by the three lines are:

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Similarly, the change in polar coordinates is:

  • x = rcosθ,
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where;

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Therefore;

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Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

brainly.com/question/14393123

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If I have a garden that's ten yards and I want to plant a plant that requires 2 yards how many plants can I plant
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Here's a method that takes advantage of the values of these particular numbers.

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... = 80×327 - 4×327

Repeated doubling will give us values that are 2, 4, and 8 times 327.

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... 2×654 = 4×327 = 654+654 = 1308 . . . . we'll use this later

... 2×1308 = 8×327 = 2616

We want 80×327, so we can add a zero to the end of this last:

... 80×327 = 26160

Now, we can subtract 4×327 to get 76×327

... 80×327 - 4×327 = 26160 -1308 = 24852 = 76×327

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More conventionally, you would multiply every digit of one number by every digit of the other and add the products according to their respective place values.

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Note the pattern of partial products here. This is a method taught to/by practitioners of Vedic mathematics, and can be done in your head. At most, you would write down the partial product sums 21, 32, 61, and 42 to keep from having to carry more than one number in your head at a time.

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