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sukhopar [10]
3 years ago
14

Given 3x = 4y = 12z, show that z = xy/(x + y) .

Mathematics
1 answer:
Alexandra [31]3 years ago
3 0
3x=4y=12z \Rightarrow x=4z,y=3z
\\ \frac{xy}{x+y} 
\\
\\ \frac{4z3z}{4z+3z}
\\
\\ \frac{12z^2}{7z}  
\\
\\ \frac{12}{7} z

So, I proved that \frac{xy}{x+y} \ne z
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PLEASE HELP! I will mark as brainliest if you help me. Solve for x. −ax + 2b > 8
anyanavicka [17]

Answer:

x < ( 8 - 2b ) / -a

Step-by-step explanation:

For this problem, we simply will solve the inequality for x.

-ax + 2b > 8

-ax > 8 - 2b

x < ( 8 - 2b ) / -a

Note, the comparison is flipped since a negative multiplicand was used to isolate x.

Cheers.

4 0
3 years ago
- Find the slope-intercept form of the equation of the line that passes through (-5, 3) and is parallel to 12x – 3y = 10.
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Use the figure to complete the proportion.
monitta

Answer:

Pretty sure it's SQ.

Step-by-step explanation:

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Can anyone do it I'll give you brainiest
Temka [501]

Answer:

<h2>64°</h2><h2 />

Step-by-step explanation:

? + 86 + 90 + 84 + 36 = 360

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Read 2 more answers
PLEASE HELP GUYS i am struggling so much, two questions
ankoles [38]

Answer: the equation of the standard parabola

1) (y-6)^2 = 4 (x-1)

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

<u>Explanation </u>

<u>Parabola:-</u>

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

  • The locus of P such that \frac{SP}{PM} = constant
  • The fixed point  'S' is called the Focus.
  • The fixed line'l 'is called the directrix of the conic
  • The constant ratio is known as the eccentricity, denoted by 'e'
  • If e=1 , the conic is called a parabola

1) <u> Step 1</u> :-

Given the focus   S = (2,6) and directrix is x=0

we know that \frac{SP}{PM}=1

now cross multiplication , we get

SP = PM

squaring on both sides,we get

SP^{2} = PM^2

step 2:-

now using distance formula is

  • \sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)

Given S =(2,6) and P(x,y) be any point on parabola

SP^2 = (x-2)^2+(y-6)^2........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

  • \frac{ax_{1}+by_{1}+c   }{\sqrt{a^2+b^2} }

Given the directrix is x =0 and P(x,y) be any point on parabola

PM^2 = \frac{x^2}{\sqrt{1}^2 }......(2)

equating equation(1) and equation (2), on simplification

we get (x-2)^2+(y-6)^2 = x^2.....(3)

  • apply (a-b)^2 = a^2+  b^2+2 ab

now the equation (3) is

(y-6)^2 = 4 x-4

now the standard form of parabola is

(y-k)^2 = 4 a(x-h)

<u>Final answer</u>:-

(y-6)^2 = 4 (x-1)

2) <u> Explanation:-</u>

<u>step 1:</u>

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

  • Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

(x-h)^2 = 4 a (y-k)

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of AS = \sqrt{(-5-(-5)^2+(2-6)^2}

 on simplification we get a =4

<u>Final answe</u>r :-

the vertex (h,k) = (-5,2) and a=4

(x-h)^2 = 4 a (y-k)

The standard parabola is (x+5)^2 = 16 (y-2)

5 0
3 years ago
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