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guajiro [1.7K]
3 years ago
5

Convert: 5 meters 10 centimeters centimeters

Mathematics
1 answer:
navik [9.2K]3 years ago
3 0
500.0000 should be right
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If a=under root s(s-a)(s-b)(s-c) then the value of a, when a =3 b =4c = 5 and s =a+b+c by 2
kramer

Answer:

6

Step-by-step explanation:

Here it's given that ,

\sf\red{\longrightarrow} A=\sqrt{s(s-a)(s-b)(s-c)}

Also ,

\sf\red{\longrightarrow} s =\dfrac{a+b+c}{2}

And we need to find out the value of A , when

  • a = 3
  • b = 4
  • c = 5

So , on substituting the respective values to find s we have ,

\sf\red{\longrightarrow} s =\dfrac{3+4+5}{2}=\dfrac{12}{2}=\bf{6}

Now let's find out A as ,

\sf\red{\longrightarrow} A = \sqrt{6(6-3)(6-4)(6-5)} \\

\sf\red{\longrightarrow}A =\sqrt{6 * 3 * 2 * 1}\\

\sf\red{\longrightarrow} A = \sqrt{3^2 * 2^2 *1^2}\\

\sf\red{\longrightarrow}A = 3*2\\

\sf\red{\longrightarrow}\boxed{\qquad \blue{\bf A = 6 \qquad}}

7 0
2 years ago
Q25 A train travelling at a speed of 30km/h crosses a bridge 25m long in 12 sec. How long will it take to overtake a another tra
Makovka662 [10]

Answer:

WHAT

Step-by-step explanation:

8 0
3 years ago
Which statement is true regarding the graphed functions?
padilas [110]

Answer:

C

Step-by-step explanation:

You can look them up in the graphs.

f(2) means "where is the blue graph when x=2?", and you can see it is at y=0.

4 0
3 years ago
Read 2 more answers
What is the correct way to write 9 x 9 x 9 in exponential form?
Harlamova29_29 [7]

Hey!

----------------------------------------------

Solution:

Put all the answer choices into expanded form.

3^9 = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3

1^8 = 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1

9^3 = 9 x 9 x 9

8^1 = 8

----------------------------------------------

Answer:

We can see that the only way to write to write 9 x 9 x 9 in exponential form is c) 9^3

----------------------------------------------

Hope This Helped! Good Luck!

5 0
3 years ago
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Exponent that will equal 32768
solmaris [256]
27 I think or something close to that hope that helps
8 0
3 years ago
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