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Tamiku [17]
2 years ago
9

Dad is twice as old as Jon. Grand-Pa is seven years older than three times Jon's age. The sum of their age is 145 years. How old

is Grand-Pa?
pls help i will give the brinlist:0
Mathematics
1 answer:
Alenkasestr [34]2 years ago
3 0

Step-by-step explanation:

let jon age be x

dad age is 2 of x i.e 2x

grandpa age is 7 + 3(x)

The sum of jon dad and grandpa age is 145 i.e

x + 2x + (7+3x) = 145

3x +. 7 + 3x = 145

6x = 145- 7

6x = 138

divide both sides by 6

x = 23

jon age is 23

while

grandpa age is

7 + 3(x)

7 + 3(23)

7 + 69

76.

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kodGreya [7K]

Answer:

(1) g[f(x)]=\frac{8x-1}{12x-4}

(2)  g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

Step-by-step explanation:

Given functions f(x) = (4x-1)

are                     g(x)=\frac{2x+1}{3x-1}

(1)  g[f(x)]=\frac{2(4x-1)+1}{3(4x-1)-1}

                =\frac{8x-2+1}{12x-3-1}

                =\frac{8x-1}{12x-4}

(2) for g^{-1}(x), rewrite the function g(x) in terms of an equation

y=\frac{2x+1}{3x-1}

Substitute y in place of x and x in place of y, then solve for y.

x=\frac{2y+1}{3y-1}

(3y-1)x = 2y+1

3xy - x = 2y + 1

3xy - 2y = x + 1

y(3x-2) = x + 1

y=(\frac{x+1}{3x-2} )

⇒ g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    f[g(x)]=(x-1)

       f[g(x)]=4[\frac{2x+1}{3x-1}] =(x-1)

⇒    \frac{(8x+4)-(3x-1)}{3x-1}=(x-1)

⇒    \frac{8x-3x+4+1}{3x-1}=(x-1)

⇒    \frac{5x+5}{(3x-1)}=(x-1)

⇒    5x+5=(3x-1)(x-1)

⇒    5x+5=3x^2-3x-x+1

⇒   5x+5=3x^2-4x+1

⇒   3x^2-9x-4=0

⇒    x=\frac{9\pm\sqrt{(-9)^2-4(-4)\times3} }{2\times3}

   x=\frac{9\pm\sqrt{81+48} }{6}

   x=\frac{9\pm\sqrt{129} }{6}

   x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

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What are the next two terms for the following sequence 1 , 8 , 27 , 64
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Answer:

125 and 216

Step-by-step explanation:

Here,

a₁ = (1)³ = 1

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Law of cosines :

The law of cosines establishes:

c ^ 2 = a ^ 2 + b ^ 2 - 2*a*b*cosC.

general guidelines:

The law of cosines is used to find the missing parts of an oblique triangle (not rectangle) when either the two-sided measurements and the included angle measure are known (SAS) or the lengths of the three sides (SSS) are known.


Law of the sines:


In ΔABC is an oblique triangle with sides a, b, and c, then:

\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}

The law of the sines is the relation between the sides and angles of triangles not rectangles (obliques). It simply states that the ratio of the length of one side of a triangle to the sine of the angle opposite to that side is equal for all sides and angles in a given triangle.

General guidelines:

To use the law of the sines you need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an opposite angle of one of them (SSA).


The ambiguous case :


If two sides and an angle opposite one of them is given, three possibilities may occur.


(1) The triangle does not exist.


(2) Two different triangles exist.


(3) Exactly a triangle exists.


If we are given two sides and an included angle of a triangle or if we are given 3 sides of a triangle, we can not use the law of the sines because we can not establish any proportion where sufficient information is known. In these two cases we must use the law of cosines

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3 years ago
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