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snow_tiger [21]
3 years ago
11

What is the LCM of 13/16a^3b^3, 11/30a^5b

Mathematics
1 answer:
vichka [17]3 years ago
8 0

Answer:

The L.C.M of given numbers is \frac{143}{240}a^5b^3.

Step-by-step explanation:

The given expressions are

\frac{13}{16}a^3b^3

\frac{11}{30}a^5b

The factors of given expressions are

\frac{13}{16}a^3b^3=\frac{13}{8}\times \frac{1}{2}\times a\times a\times a\times b\times b\times b

\frac{11}{30}a^5b=\frac{11}{15}\times \frac{1}{2}\times a\times a\times a\times a\times a\times b

L.C.M.=\frac{13}{8}\times \frac{1}{2}\times a\times a\times a\times b\times b\times b\times \frac{11}{15}\times a\times a

L.C.M.=\frac{143}{240}\times a^5\times b^3

L.C.M.=\frac{143}{240}a^5b^3

Therefore, the L.C.M of given numbers is \frac{143}{240}a^5b^3.


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A dolphin jumps from the water at an initial velocity of 16 feet per second. The equation h + -16t^2 + 80t models the dolphin's
Dominik [7]

Answer:

100feet

Step-by-step explanation:

Given the height reached by a dolphin expressed by the equation is h = -16t^2 + 80t. At maximum height, the velocity is zero, hence;

v(t) = dh/dt = -32t + 80

0 = -32t + 80

32t = 80

t = 80/32

t = 2.5secs

Substitute t = 2.5 into the height formula h = -16t^2 + 80t

h = -16(2.5)^2 + 80(2.5)

h = -100+200

h = 100feet

Hence  the maximum height the dolphin jumps is 100feet

3 0
2 years ago
Need to know the area of the composite figure.
Deffense [45]

Answer:

  51 m^2

Step-by-step explanation:

The shaded area is the difference between the area of the overall figure and that of the rectangular cutout.

The applicable formulas are ...

  area of a triangle:

     A = (1/2)bh

  area of a rectangle:

     A = bh

  area of a trapezoid:

     A = (1/2)(b1 +b2)h

___

We note that the area of a triangle depends only on the length of its base and its height. The actual shape does not matter. Thus, we can shift the peak of the triangular portion of the shape (that portion above the top horizontal line) so that it lines up with one vertical side or the other of the figure. That makes the overall shape a trapezoid with bases 16 m and 10 m. The area of that trapezoid is then ...

  A = (1/2)(16 m + 10 m)(5 m) = 65 m^2

The area of the white internal rectangle is ...

  A = (2 m)(7 m) = 14 m^2

So, the shaded area is the difference:

  65 m^2 -14 m^2 = 51 m^2 . . . . shaded area of the composite figure

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<em>Alternate approach</em>

Of course, you can also figure the area by adding the area of the triangular "roof" to the area of the larger rectangle, then subtracting the area of the smaller rectangle. Using the above formulas, that approach gives ...

  (1/2)(5 m)(16 m - 10 m) + (5 m)(10 m) - (2 m)(7 m) = 15 m^2 + 50 m^2 -14 m^2

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7 0
3 years ago
A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr
seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

Perimeter = a\sqrt{2} + b\sqrt{10}

Required

a + b

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

(x_1,y_1) = A(0,1)

(x_2,y_2) = B(3,4)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}

AB = \sqrt{( - 3)^2 + (-3)^2}

AB = \sqrt{9+ 9}

AB = \sqrt{18}

AB = \sqrt{9*2}

AB = \sqrt{9}*\sqrt{2}

AB = 3\sqrt{2}

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

(x_1,y_1) = B (3,4)

(x_2,y_2) = C(4,3)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}

BC = \sqrt{(-1)^2 + (1)^2}

BC = \sqrt{1 + 1}

BC = \sqrt{2}

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = C(4,3)

(x_2,y_2) = D (3,0)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}

CD = \sqrt{(1)^2 + (3)^2}

CD = \sqrt{1 + 9}

CD = \sqrt{10}

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = D (3,0)

(x_2,y_2) = A (0,1)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}

DA = \sqrt{(3)^2 + (- 1)^2}

DA = \sqrt{9 +  1}

DA = \sqrt{10}

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}

Perimeter = 4\sqrt{2} + 2\sqrt{10}

Recall that

Perimeter = a\sqrt{2} + b\sqrt{10}

This implies that

a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}

By comparison

a\sqrt{2} = 4\sqrt{2}

Divide both sides by \sqrt{2}

a = 4

By comparison

b\sqrt{10} = 2\sqrt{10}

Divide both sides by \sqrt{10}

b = 2

Hence,

a + b = 2 + 10

a + b = 12

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