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2 years ago

Rewrite 5ab showing the multiplication signs

Mathematics

2 answers:

monitta2 years ago

8 0

Answer:

$5ab = 5 \times a \times b$

Step-by-step explanation:

horrorfan [7]2 years ago

7 0

Answer:

Step-by-step explanation:

5a*b

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A man is five times as old as his son. Four years ago the man was thirteen times as old as his son.

skad [1K]

Answer:

Age of son = 6 years

Age of man = 5×6 = 30 years

Step-by-step explanation:

GIVEN :-

A man is 5 times as old as his son. (In Present)
4 years ago , the man was 13 times as old as his son

TO FIND :-

The present ages of the man & his son.

SOLUTION :-

Let the present age of son be 'x'.

⇒ Present age of man = 5x

4 years ago ,

Age of son = (Present age of son) - 4 = x - 4

Age of man = (Present age of man) - 4 = 5x - 4

The man was thirteen times as old as his son. So,

$=> 5x - 4 = 13(x - 4)$

Now , solve the equation.

Open the brackets in R.H.S.

$=> 5x - 4 = 13x - 52$

Take 5x to R.H.S. and -52 to L.H.S. Also , take care of their signs because they are getting displaced from L.H.S. to R.H.S. or vice-versa.

$=> 13x - 5x = 52 - 4$

$=> 8x = 48$

Divide both the sides by 8

$=> \frac{8x}{8} = \frac{48}{8}$

$=> x = 6$

CONCLUSION :-

Age of son = 6 years

Age of man = 5×6 = 30 years

8 0

2 years ago

Convert to a decimal. Be sure to mark repeating decimals.

Sergio [31]

Answer:

$.24\\is\\reapeating\\0.24$

Step-by-step explanation:

8 0

2 years ago

To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)

Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}$

$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$