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3 years ago

What is the u known decimal equivalent for 5/100.

Mathematics

2 answers:

Vadim26 [7]3 years ago

8 0

0.05
So 5÷100=0.05

0.05÷5=100

dexar [7]3 years ago

7 0

I think that answer would be .05

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Factor 7d+56<br> I'll give Brainless.....

Lesechka [4]

Answer: 7(d+56)=63d

Step-by-step explanation:

U add the 7 +56 and the answer is 63d plz mark brainly

7 0

3 years ago

Deon is riding his bicycle. He rides for 7 hours at a speed of 22.4 kilometers per hour. For how many kilometers does he ride

salantis [7]

By hypothesis, Deon is riding at a speed of 22.4 kilometers per hour. So each hour, he completes 22.4 km.

If he rides for 7 hours, he completes 22.4 * 7 = 156.8 km

So in 7 hours, he rides 156.8 kilometers.

Hope this helps! :)

6 0

4 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

In this line of circles, the ratio of red circles to blue circles is 1:3 How many more circles would need to be added to make th

zaharov [31]

Answer:

The number of red circles to be added would be equal to 8 times the original amount of red circles

Step-by-step explanation:

Let

x ----> the number of red circles

y ----> the number of blue circles

Originally

$\frac{x}{y}=\frac{1}{3}$

$y=3x$ ----> equation A

How many more circles would need to be added to make the ratio of red to blue 3:1

Let

n ----> the number of additional red circles

$\frac{x+n}{y}=\frac{3}{1}$ ----> equation B

substitute equation A in equation B

$\frac{x+n}{3x}=\frac{3}{1}$

solve for n

$x+n=9x\\n=9x-x\\n=8x$

therefore

The number of red circles to be added would be equal to 8 times the original amount of red circles

5 0

3 years ago

Simplify the expression where possible. (6^3)^-3

astraxan [27]

$\bf ~\hspace{7em}\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
~\hspace{4.5em}
a^n\implies \cfrac{1}{a^{-n}}
~\hspace{4.5em}
\cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
(6^3)^{-3}\implies 6^{-3\cdot 3}\implies 6^{-9}\implies \cfrac{1}{6^9}\implies \cfrac{1}{10077696}$

3 0

3 years ago

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