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Olenka [21]
3 years ago
5

Solve the proportion using cross products.

Mathematics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

  h = 20

Step-by-step explanation:

The idea of a "cross product" is that you eliminate denominators by multiplying each numerator by the denominator on the other side of the equality. This is what it looks like:

  \dfrac{29\,\text{miles}}{58\,\text{hours}}=\dfrac{10\,\text{miles}}{h\,\text{hours}}\ \longrightarrow\ (29\,\text{miles})(h\,\text{hours})=(10\,\text{miles})(58\,\text{hours})

Now, you can divide by the units and the coefficient of h to find ...

  h=\dfrac{10\cdot 58}{29}=20

_____

<em>Comment on cross product</em>

IMO, you need to go into the idea of "cross product" with eyes wide open. The rules of equality require that any operation performed on one side of the equation also be performed on the other side. The notion of "cross product" <em>makes it look like you're doing something different</em> to the two sides of the equation.

In fact, the "cross product" is a shortcut for "multiply both sides of the equation by the product of the denominators." Since each denominator term cancels itself, multiplying by the product of denominators <em>looks like</em> multiplying by the other denominator (and throwing away the denominator you have).

__

<em>Another comment on cross product</em>

The idea of "using cross products" is often suggested wherever one fraction is equal to another. In many cases, that is an extra step that is unnecessary.

If the variable is in the numerator, you need only multiply the equation by its denominator, not both of them. If you multiply by both denominators, you find yourself dividing again by the denominator of the fraction not containing the variable. For example, ...

  h/10 = 58/29

using cross products, this is ...

  29h = 10·58

and we need to divide by the 29 that we just multiplied by.

Instead, we could simply multiply by the denominator of h:

  h/10 = 58/29   ⇒   h = 10·58/29

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the perimeter of a basketball court is 102 meters, and length is 6 meters longer than twice the width, what r length and width.
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Perimeter = 2(l + w)

Thus,

102 = 2(l + w)

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Thus, w = 15.

Length = 6 + 2w
= 6 + 30
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Thus, the length is 36 meters and width is 15 meters.
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Write 9 decimals with one decimal place that when rounded to the nearest one round to 7
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Y+7=-2(x-1)<br><br><br> hurry plz
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y=-2x-5

Step-by-step explanation:

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8 0
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Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4
dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

x_1\ =\ -2

y_1\ =\ 3

z_1\ =\ -4

Distance is measure across the line

\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}

So, we can write

\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k

=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k

=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k

=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3

=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18

=>18k-24+12k+9+10k-32\ =\ 18

=>\ k\ =\dfrac{13}{8}

So,

x\ =\ 3k-4

   =\ 3\times \dfrac{13}{8}-4

   =\ \dfrac{7}{4}

y\ =\ \dfrac{4k+3}{2}

   =\ \dfrac{4\times \dfrac{13}{8}+3}{2}

   =\ \dfrac{19}{4}

z\ =\ \dfrac{5k-16}{3}

  =\ \dfrac{5\times \dfrac{13}{8}-16}{3}

   =\ \dfrac{-21}{8}

Now, the distance of point from the plane is given by,

d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}

   =\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}

   =\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}

   =\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}

   =\ \sqrt{\dfrac{1177}{64}}

   =\ 4.28

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

            = 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0
4 years ago
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