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3 years ago

Someone help me with these math questions

Mathematics

2 answers:

Maslowich3 years ago

8 0

48 mph
Hope this helps :)

mart [117]3 years ago

3 0

Well for the first one the ratio would be 12:6 I believe

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Given that Flx) = x2 + 2, evaluate F(2).

wariber [46]

Step-by-step explanation:

f(2) = 2^2 +2

= 4+2

3 0

3 years ago

A store is instructed by corporate headquarters to put a markup of 25% on all items. An item costing $24 is displayed by the s

larisa [96]

Answer: The correct selling price is $29.97.

Step-by-step explanation:

Since we have given that

Cost price of an item = $27

Mark up rate = 11%

So, Amount of mark up would be

So, Amount after mark up would be

Hence, the correct selling price is $29.97.

The manager's likely error is that he has put the selling price the mark up amount only i.e $2.9≈$3 instead of adding the mark up amount to the cost price.

5 0

3 years ago

Read 2 more answers

Help!!! (Tell me if you can’t see it clearly)

meriva

I believe it’s option c!

3 0

3 years ago

Please help me ASAP only answer if you’re absolutely sure and/or you took the benchmark

ICE Princess25 [194]

Answer: the answer is a

Step-by-step explanation:

8 0

3 years ago

Which expression is equivalent to *picture attached*

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;

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

4 0

3 years ago