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Andrews [41]
2 years ago
11

what are the solutions to this polynominal when s(x) =0 s(x) (3x^2 + 17x+10 )(x-7)(4x-3) enter your answers in the boxes, from l

east to greatest.
Mathematics
2 answers:
Juli2301 [7.4K]2 years ago
8 0

Answer:

-5, -2/3, 3/4, 7

Step-by-step explanation:

The solutions are solutions of one of the following:

(3x^2 + 17x + 10) = 0, x - 7 = 0, 4x - 3 = 0

The latter two are straightforward: x = 7 or 3/4.

To factor the first, note it must be of the form (3x + __) (x + __)

Check the pairs (10, 1) (1, 10) (5, 2) and (2, 5) and we find:

(3x + 2) (x + 5) = 0

So, we also have solutions -5, -2/3

SOVA2 [1]2 years ago
4 0

Answer:-5,-2/3,3/4,7

Step-by-step explanation:

took quiz

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There are 16 players in my poker group that meets the first Tuesday of every month. Ten of the players are male and six are fema
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The number of picces of art at the musem is shown in the table.
densk [106]

Answer:

<u>The solution is multiple:</u>

<u>A. 15 artists with 2 paintings per artist.</u>

<u>B. 10 artists with 3 painting per artist.</u>

<u>C. 6 artists with 5 painting per artist.</u>

<u>D. 5 artists with 6 paintings per artist.</u>

<u>E. 3 artists with 10 paintings per artist.</u>

<u>F. 2 artists with 15 paintings per artist.</u>

Statement and question complete:

The number of pieces of art at a museum is shown in the table.

Oil paintings 30

Photographs 24

Sketches 21.

The museum is hosting a show for July that features the oil paintings and each artist will show more than 1 painting. How many artists could be featured in the show?

Source: https://educationexpert.net/mathematics/929012.html

Step-by-step explanation:

x = Number of paintings per artist

y = Total number of paintings

x ≥ 2

xy = 30

<u>The solution is multiple:</u>

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<u>B. 10 artists with 3 painting per artist.</u>

<u>C. 6 artists with 5 painting per artist.</u>

<u>D. 5 artists with 6 paintings per artist.</u>

<u>E. 3 artists with 10 paintings per artist.</u>

<u>F. 2 artists with 15 paintings per artist.</u>

8 0
3 years ago
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