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nexus9112 [7]
2 years ago
15

Simon Simms takes out $20,000 of a twenty-payment life when he is twenty-five years old. a. annual premium b. quarterly premium

Mathematics
1 answer:
OLga [1]2 years ago
6 0
B. Quarterly premium
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The following points are dilated at a center at point C by each of the following scale factors.
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Answer:

(a) A is dilated by a scale factor of 3.

Step-by-step explanation:

where the imagine

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A student has four sticks of wood that measures 3 inches, 4 inches, 5 inches, and 6 inches.Can the student use any three of thes
saw5 [17]

Answer:

Yes, the student can use three of these sticks to make a right triangle.

Step-by-step explanation:

The measure of four sticks = 3 inches, 4 inches, 5 inches and 6 inches.

Now, in a right angled triangle, by PYTHAGORAS THEOREM the three sides of the triangle base, perpendicular and  hypotenuse are in the  given following relation :

(Base)^{2}  + (Perpendicular)^{2}   = (Hypotenuse)^{2}

Now, here among all four sticks

(3)^{2}  +(4)^{2}   = 9 + 16  = 25

and 25 = (5)^{2}

So, there exists three sticks of length 3 in, 4 in and 5 inches which satisft Pythagoras Theorem , such that:

(3)^{2}  + (4)^{2}   = (5)^{2}

So, the required length sticks are 3 inches, 4 inches and 5 inches.

Hence, the student use three of these sticks to make a right triangle.

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3 years ago
Drew dis its one of kmk charactersssss
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Calculate the product: 45.67 X 32.58
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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
2 years ago
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