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3 years ago

(Brainliest for fastest right answer!!!) Which is correct?

Mathematics

1 answer:

zavuch27 [327]3 years ago

5 0

Answer:

Step-by-step explanation:

ax² + bx + c = a(x - $x_{1}$ )(x - $x_{2}$ )

D = b² - 4ac

$x_{12}$ = ( - b ± √D ) / 2a

~~~~~~~~~~~~

Let x² = y , then y² + 8y - 9

Find the roots of equation y² + 8y - 9 = 0

D = 64 + 36 = 100 = 10²

$y_{1}$ = ( - 8 - 10) / 2 = - 9

$y_{2}$ = ( - 8 + 10) / 2 = 1

y² + 8y - 9 = (y - 1)(y + 9) = (x² - 1)(x² + 9)

x² = 1 ⇒ $x_{12}$ = ± √1

$x^{4}$ + 8x² - 9 = (x - 1)(x + 1)(x² + 9)

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On each of n plates there are k apples and m pears. What is the total number of fruit are there on these plates?

Tatiana [17]

Answer:

I Have noidea

Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

what is the answer here 90 student went to the zoo, 3 had hamburger milk and cake; had 5 milk and hamburger; 10 had cake and mil

dangina [55]

Answer:

a) 37

b) 2

c) 17

d) 8

Step-by-step explanation:

90 Students went to the zoo. 3 had hamburger, milk and cake; 5 had milk and hamburger, 10 had cake and milk; 8 had cake and hamburger; 24 had hamburger; 38 had cake; 20 had milk. How many had a. nothing b. cake only c. milk only d. hamburger only

Solution:

Let h represent students that ate hamburger, m represent students that had milk and c represent students that had cake.

Given that:

n(h ∩ m ∩ c) = 3, n(m ∩ h) = 5, n(c ∩ m) = 10, n(c ∩ h) = 8, n(h) = 24, n(c) = 38, n(m) = 20

The number of students that had nothing = n(h ∪ m ∪ C)'

The number of students that had only milk = n(m ∩ h' ∩ C')

The number of students that had only cake = n(m' ∩ h' ∩ C)

The number of students that had only hamburger = n(m' ∩ h ∩ C')

a) n(m ∩ h' ∩ C') = n(m) - n(m ∩ h) - n(c ∩ m) - n(h ∩ m ∩ c) = 20 - 5 - 10 - 3 = 2

n(m' ∩ h ∩ C') = n(h) - n(m ∩ h) - n(c ∩ h) - n(h ∩ m ∩ c) = 24 - 5 - 8 - 3 = 8

n(m' ∩ h' ∩ C) = n(m) - n(m ∩ c) - n(c ∩ h) - n(h ∩ m ∩ c) = 38 - 10 - 8 - 3 = 17

n(m ∩ h' ∩ C') + n(m' ∩ h ∩ C') + n(m' ∩ h' ∩ C) + n(h ∪ m ∪ C)' + n(h ∩ m ∩ c) + n(m ∩ h) + n(c ∩ m) + n(c ∩ h) = 90

2 + 8 + 17 + 5 + 10 + 8 + 3 + n(h ∪ m ∪ C)' = 90

53 + n(h ∪ m ∪ C)' = 90

n(h ∪ m ∪ C)' = 37

b) n(m' ∩ h' ∩ C) = 17

c) n(m ∩ h' ∩ C') = 2

d) n(m' ∩ h ∩ C') = 8

4 0

3 years ago

According to the data, the mean quantitative score on a standardized test for female college-bound high school seniors was 500.

Sveta_85 [38]

Answer:

6.68% of the female college-bound high school seniors had scores above 575.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(scores above 575)

P(x > 575)

$P( x > 575) = P( z > \displaystyle\frac{575 - 500}{50}) = P(z > 1.5)$

$= 1 - P(z \leq 1.5)$

Calculation the value from standard normal z table, we have,

$P(x > 575) = 1 - 0.9332= 0.0668 = 6.68\%$

6.68% of the female college-bound high school seniors had scores above 575.

7 0

3 years ago

Recall, we have five connectives in propositional logic ¬, ∧, ∨, →, ↔ (negation, conjunction, disjunction, conditional and bicon

Free_Kalibri [48]

Answer:

(a) ¬(p→¬q)

(b) ¬p→q

Step-by-step explanation

taking into account the truth table for the conditional connective:

p | q | p→q

T | T | T

T | F | F

F | T | T

F | F | T

(a) and (b) can be seen from truth tables:

for (a) p∧q:

p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q

T | T | F | F | T | T

T | F | T | T | F | F

F | T | F | T | F | F

F | F | T | T | F | F

As they have the same truth table, they are equivalent.

In a similar manner, for (b) p∨q:

p | q | ¬p | ¬p→q | p∨q

T | T | F | T | T

T | F | F | T | T

F | T | T | T | T

F | F | T | F | F

again, the truth tables are the same.

For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))

4 0

3 years ago

Pls help me I'll give brainlest and points

dmitriy555 [2]

Answer:

See below.

Step-by-step explanation:

O is the origin, so its abscissa is 0. O: 0

The only point 4 units from O is A, so A: 4

Point C has a positive abscissa like A.

Points B and C have opposite abscissas. The only opposites are -3 and 3, but C is positive, and B is negative.

B: -3

C: 3

D: -4.5

E: -6

M = (4 + (-6))/2 = -1. M: -1

OB = |0 - (-3)| = |3| = 3

DA = |-4.5 - 4| = |-8.5| = 8.5

4 0

3 years ago