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3 years ago

Give the domain and range.

Mathematics

2 answers:

3241004551 [841]3 years ago

8 0

Answer:

The correct answer is a.

Step-by-step explanation:

The domain is the plotted points on the x-axis.

The range is the plotted points on the y-axis.

(-3, 3), (0, 0), (2, -2)

Domain {-3, 0, 2}; Range: {3, 0, -2}

shutvik [7]3 years ago

5 0

Answer:Its A

Step-by-step explanation:Because it is ^_^

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Whats the area of 8ft, 6ft and 2ft?

Vitek1552 [10]

Answer:

Step-by-step explanation:

8x6=48

48x2=96

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3 years ago

sophia borrowed $12,109 at 4.5% simple interest for 6 months. How much will the interest amount to? What is the total amount tha

Yuri [45]

The applicable formula is
I = Prt
where P = the principal value (12109), r is the interest rate (0.045), and t is the time period in years (1/2)

a) Substituting the given values, the interest amount is computed as
I = ($12,109)(.0.45)(1/2)
I = $272.45

b) The amount Sophia will have to pay back is the sum of the principal amount and the interest owed.
$12,109 + 272.45 = $12,381.45

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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;

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

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</span>

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3 years ago

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