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ale4655 [162]
2 years ago
15

24. If all of the dimensions of the aquarium

Mathematics
1 answer:
Zigmanuir [339]2 years ago
8 0

Answer:

You can solve this problem by doubling each original measure then putting them in the equation. So, a length of 6in becomes 12in, a width of 10in becomes 20in, a height of 9in becomes 18in. The equation is now 12x20x18. 12x20= 240. 240x18= 4,320. This means the new aquarium has a volume of 4,320in3.

Step-by-step explanation:

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-25x^2+xy-21y^2+6x+8
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X^2+16x+64 can anyone show the work for this and find the factor.
Alexandra [31]

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should i simplify it? or factorise it?

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x^2+16+64

(x)^2+2×x×8+(8)^2

(x+8)^2

=(x+8)(x+8)

8 0
3 years ago
A can of chili has a radius of 5.25 cm and a height of 13 cm. find the volume
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Step-by-step explanation:

4 0
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Given the following diagram, find the required measure.
Zielflug [23.3K]

Answer:

∠ 5 = 40°

Step-by-step explanation:

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3 0
3 years ago
find the parametric equations for the line of intersection of the two planes z = x + y and 5x - y + 2z = 2. Use your equations t
Kaylis [27]

Answer:

You didn't give the points in which you want the parametric equations be filled, but I have obtained the parametric equations, and they are:

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

Step-by-step explanation:

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

r = r_0 + tv

where r_0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

x = ax, y = by, and z = cz

where a, b and c are the coefficients from the vector equation

r = ai + bj + ck

To find the parametric equations for the line of intersection of the planes.

x + y - z = 0

5x - y + 2z = 2

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find v, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are:

For the plane x + y - z = 0, the normal vector is a⟨1, 1, -1⟩

For the plane 5x - y + 2z = 2, the normal vector is b⟨5, -1, 2⟩

The cross product of the normal vectors is

v = a × b =

|i j k|

|1 1 -1|

|5 -1 2|

= i(2 - 1) - j(2 + 5) + k(-1 - 5)

= i - 7j - 6k

v = ⟨1, -7, -6⟩

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set z = 0 in both equations, we get

x + y = 0

5x - y = 2

Adding these equations

5x + x + y - y = 2 + 0

6x = 2

x = 1/3

Substituting x = 1/3 back into

x + y = 0

y = -1/3

Putting these values together, the point on the line of intersection is

(1/3, -1/3, 0)

r_0= (1/3) i - (1/3) j + 0 k

r_0​​ = ⟨1/3, -1/3, 0⟩

Now we’ll plug v and r_0​​ into the vector equation.

r = r_0​​ + tv

r = (1/3)i - (1/3)j + 0k + t(i - 7j - 6k)

= (1/3 + t) i - (1/3 + 7t) j - 6t k

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up r = ai + bj + ck with our vector equation,

r = (1/3 + t) i + (-1/3 - 7t) j + (-6t) k

a = (1/3 + t)

b = (-1/3 - 7t)

c = -6t

Therefore, the parametric equations for the line of intersection are

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

3 0
3 years ago
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