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aliina [53]
2 years ago
5

Calculate 13.5 % of $ 660

Mathematics
1 answer:
atroni [7]2 years ago
8 0

Answer:

$89.1

Step-by-step explanation:

13.5 % of $660

= 13.5/ 100 × 660

= 0.135 × 660

= 89.1

hence, $ 89.1

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Use Euler's formula to find the missing number.
frutty [35]

Answer:

15

Step-by-step explanation:

trust me bro

6 0
3 years ago
The sum of two numbers is 12. The difference of the same two numbers is 24. what is the larger of the two numbers
Vsevolod [243]

ANSWER


The larger number is 18

<u>EXPLANATION</u>

Let x be the larger number and y be the smaller  number.


Then,

x+y=12--(1)


x-y=24--(2)


Let us add equation (1) and (2).


2x=36


Divide through by 2


x=18


Put x=18 in to equation (1).

This implies that,

18=y=12



y=12-18=-6




4 0
3 years ago
483.6 is what percent of 180
liraira [26]

Answer:

We assume, that the number 180 is 100% - because it's the output value of the task.

2. We assume, that x is the value we are looking for.

3. If 100% equals 180, so we can write it down as 100%=180.

4. We know, that x% equals 483.6 of the output value, so we can write it down as x%=483.6.

5. Now we have two simple equations:

1) 100%=180

2) x%=483.6

where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:

100%/x%=180/483.6

6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for 483.6 is what percent of 180

100%/x%=180/483.6

(100/x)*x=(180/483.6)*x       - we multiply both sides of the equation by x

100=0.37220843672457*x       - we divide both sides of the equation by (0.37220843672457) to get x

100/0.37220843672457=x

268.66666666667=x

x=268.66666666667

now we have:

483.6 is 268.66666666667% of 180

4 0
3 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
Help plssssssssssssss
morpeh [17]

Answer:

£307.80

Hope this helps!

3 0
2 years ago
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