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ki77a [65]
3 years ago
6

If f(x) varies directly with x and f(x) = 72 when x = 6, find the value of f(x) when x = 3.

Mathematics
2 answers:
mestny [16]3 years ago
7 0

Answer:

The value of f(x)  at x = 6 is 36

Step-by-step explanation:

Given as :

f(x) varies directly with x

f(x) = 72  ,  for x = 6

So, The function f(x) must be 12 x

i.e f(x) = 12 x

A this function satisfy for x = 6

Now, The function is f(x) = 12 x

So, for x = 3

The function f(x) = 12 × 3

I.e  f(x) = 36

Hence The value of f(x)  at x = 6 is 36  Answer

yaroslaw [1]3 years ago
6 0

Answer:

36

Step-by-step explanation:

Since f(x) varies directly with x, f(x) can be expressed alternatively as \[f(x) = k * x\] where k is a constant value.

Given that f(x) is 72 when the value of x is 6.

This implies, \[72 = k * 6\]

Simplifying and rearranging the equation to find the value of k:

k = \frac{72}{6}

Hence k = 12

Or, \[f(x) = 12 * x\]

When x = 3, \[f(x) = 12 *3 \]

Or in other words, the value of f(x) when x=3 is 36

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The equivalent expression of -14 - 6 is -14 - (+6)

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Equivalent expressions are expressions that have the same value when evaluated

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The expression is given as:

-14 - 6

Put the terms of the expression in brackets

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-14 - (+6)

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Type the correct answer in each box. Use numerals instead of words.
lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a = 5/4, b =0 and c = -17

x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17    

8(-2) - y = -17    

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

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