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4 years ago

Find the roots for y = 3x2 - 5x - 1

Mathematics

1 answer:

Mars2501 [29]4 years ago

8 0

Step-by-step explanation:

3x squared - 5x - 1 equals 0 using the method of completing the square he gets 3 x square - 5 x equals one and then x squared - 5/3 x is equal to 1 what does dividing should I pray then you come down you get X squared - 5/3 x into (- 5/6 or squared is equals 1/3 + but it will be negative 5/6 or squared you good like them send any you get so at the end your roots are going to be x equals 1.847 or x equals negative 0.180 results

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3 years ago

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Stella [2.4K]

Answer:

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Step-by-step explanation:

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3 years ago

23 and a half divided by 4

kkurt [141]

Answer:

$\frac{47}{8}$

Step-by-step explanation:

1) Rewrite the words into an equation:

$23\frac{1}{2}$ ÷ 4

2) Change the division sign to a multiplication sign and find the reciprocal of 4 (flip the numerator and denominator of 4):

$23\frac{1}{2} * \frac{1}{4}$

3) Change the mixed number into an improper fraction:

$\frac{47}{2} * \frac{1}{4}$

4) Multiply:

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4 years ago

Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1

Rudik [331]

Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = $N_T$

$N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960$

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

$P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}$

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are $({}^{10}C_1)$ .

For each card we can draw from any of the 4 sets. It can be described mathematically as: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$