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3 years ago

Express Express 0.45 as a percent

Mathematics

2 answers:

kkurt [141]3 years ago

5 0

It’s 45%. This is because when changing a decimal into a fraction you move the decimal over two times to the right.

mihalych1998 [28]3 years ago

4 0

To change 0.45 to a percent, we can move the decimal point 2 places to the right and add the percent sign.

If we move the decimal 2 places to the right, we will end up with 45 and then we will add the percent sign and it will become 45%

Therefore, 0.45 to a percent is 45%

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Answer:

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Step-by-step explanation:

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Answer:

Horizontal distance = 0 m and 6 m

Step-by-step explanation:

Height of a rider in a roller coaster has been defined by the equation,

y = $\frac{1}{3}x^{2}-2x+8$

Here x = rider's horizontal distance from the start of the ride

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ii). Since, the parabolic graph for the given equation opens upwards,

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From the equation,

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Therefore, minimum height of the rider will be the y-coordinate of the vertex.

Minimum height of the rider = 5 m

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Step-by-step explanation:

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DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

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