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3 years ago

Factor to find the zeros of the function defined by the quadratic expression. −13x2 − 130x − 273

Mathematics

1 answer:

rewona [7]3 years ago

7 0

Answer The Zeros are x= -3,-7

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A basketball was originally priced at $25, but Tucker waited to buy it until the basketball was on sale for 20% off. If he paid

kotegsom [21]

Answer:

The answer to your question is $21.0

Step-by-step explanation:

Data

Original price = $25

Discount = 20%

Taxes = 5%

Process

1.- Calculate the price with the discount

$25 ------------------- 100%

x ------------------- 20%

x = (20 x 25) / 100

x = 500 / 100

x = $5

Price after the discount = $25 - $5

= $20

2.- Calculate the price after taxes

$20 ---------------------- 100%

x ----------------------- 5%

x = (5 x 20) / 100

x = 100 / 100

x = $1

Final price = $20 + $1

= $21

3 0

3 years ago

Solve this the link is down below

kogti [31]

There is no link here

4 0

3 years ago

Bill Payne visits his local bank to see how long it will take for $1,000 to amount to $1,900 at a simple interest rate of 12½%.

nikklg [1K]

My life thsi is my life not your life

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

How does the numbers of zeros in the product of 8 and 5000 compare to the number of zeros in the factors?

o-na [289]

If product means to multiply than
5000*8= 40000
Answer: There is one more zero in the product.

6 0

3 years ago