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2 years ago

Solve: I -12 I - I 3 I =

Mathematics

2 answers:

Alborosie2 years ago

8 0

Answer: A

Step-by-step explanation:

-12 is made positive so the equation is turned to 12-3 which is 9

hammer [34]2 years ago

7 0

Answer:

Step-by-step explanation:

You don't need a exp so put the answer in

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Cause I lowkey suck at math How do I do this

dimaraw [331]

1. The slope is -2/5 not 3-/10

5 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

Determine the equation of the parabola shown in the diagram in factored form. (check picture below)

krok68 [10]

Answer:

$(x+1)(x-3)$

Step-by-step explanation:

A quadratic in factored form is usually expressed as: $a(x\pm a)(x \pm b)$ where the sign of a and b depends on the sign of the zero. And I said "usually" since sometimes the x will have a coefficient. Anyways in the quadratic there are two zeroes at x=-1 and x=3. This can be written as: $a(x+1)(x-3)$ . Notice how the signs are different? This is because when you plug in -1 as x you get a factor of (-1+1) which becomes 0 and it makes the entire thing zero since when you multiply by 0, you get 0. Same thing for the x-3 if you plug in x=3. Now a is in front and it can influence the stretch/compression. To find the value of a, you can take any point (except for the zeroes, because it will make the entire thing zero, and you can technically input anything in as a)