Which pair of values could NOT be the roots of a quadratic polynomial?
A) 4 and 6
B) 5 and 5
C) 2 + 3i and 2 + 3i
D) 1 + 3i and 1 –3i

Answer:

a) 3

b) 725

c) 12100

d)3900

Answer:

Step-by-step explanation:

1. If I do not grumble while doing the logic exam, then it is because I understand the questions.

2. If the arguments are not arranged in regular order, then I will not be used to these examples.

3. If my headaches, then it is probably because it is not an easy example.

4.  If the arguments are not arranged in regular order like I am used to, then I will not be able to understand them.

5. If the example is not easy and I am grumbling, then it is because it is giving me a headache.

Answer:

y = -7

Step-by-step explanation:

✔️Find the slope (m) using any two pairs of values, say (0, -7) and (2, -7):

Slope (m) = ∆y/∆x = (-7 -(-7))/(2 - 0) = 0/2 = 0

Slope (m) = 0

✔️Find y-intercept (b):

y-intercept is the value of y when x = 0. From the table, when x = 0, y = -7.

Therefore, y-intercept (b) = -7

✔️ Substitute m = 0 and b = -7 into y = mx + b (slope-intercept form):

y = 0(x) + (-7)

y = -7

Answer:

A 5/13

Step-by-step explanation:

Just took the test

The value of Cos L in the given right triangle JKL, is 5/13

The cosine of an angle is defined as the sine of the complementary angle.

The complementary angle equals the given angle subtracted from a right angle, 90°.

Also, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H).

Given is a right triangle JKL, we need to find the value of cos(L),

Since, the cosine of an angle is the ratio of the base and hypotenuse,

So, firstly of we will find the value of the hypotenuse,

Using the Pythagoras theorem,

12²+5² = hypotenuse²

hypotenuse = √(12²+5²)

hypotenuse = √169

hypotenuse = 13

Therefore, Cos L = 5/13

Hence, the value of Cos L in the given right triangle JKL, is 5/13

Learn more about cosine of the angles, click;

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Step-by-step explanation:

1. given

2. vertical angles

3. alternate interior angles

4. AA I think

Answer:

number 1 hehehehej2b2ii2

The Amount of property taxes you would have to pay per year if the rate was 0.79% is $3,436.50.

If you owned a house and the assessed value was $435,000.00, the amount of property taxes you would have to pay per year

if the rate was 0.79% can be calculated as follows:SolutionTo calculate the amount of property taxes, multiply the assessed value by the tax rate.

We can represent this mathematically as:P = R × Vwhere:P = Property taxesR = Tax rateV = Assessed Value

We are given that:R = 0.79%V = $435,000.00

We need to convert the tax rate to a decimal value before we can use it in our calculation. This can be done by dividing the percentage value by 100. Therefore:R = 0.79 ÷ 100= 0.0079

Using the formula above, we can now calculate the property taxes as:P = R × V= 0.0079 × $435,000.00= $3,436.50

Therefore, the amount of property taxes you would have to pay per year if the rate was 0.79% is $3,436.50.

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Answer:

1/3

Step-by-step explanation:

Answer:

sorry

Step-by-step explanation:

Where is the figure?

This is a simple division problem, there are 1000 grams in 1 kilogram, so we do 2000/8, which is 250, so Erin put 250 grams of soil in each pot.

Answer:

120.

Step-by-step explanation:

Since Alex and 4 friends are playing baseball on the same team, and each time they play a game, they switch the batting order to a sequence that has not yet been used, to determine what is the maximum number of games they can play before they have to repeat a batting order they have already used, the following calculation must be performed:

1 x 2 x 3 x 4 x 5 = X

2 x 3 x 4 x 5 = X

6 x 4 x 5 = X

24 x 5 = X

120 = X

Thus, the maximum number of games they can play before they have to repeat a batting order they have already used is 120.

Answer:

F(x) =2(1/2 x2-6x) +C

Step-by-step explanation:

Hope it helps

Answer: the correct answer is B. f(x)=[x]

Step-by-step explanation:

1. The parent function of a Function part full is shown below:

f(x)=[x] (This is the simplest form) Because from this function we can make transformations that allow us to obtain the function f(x) =2[x-6]

2. If you take the parent function and make y=f(x-6), then you have:

f(x-6)=[x-6] (The function is shifted 6 units on the positive x-axis)

3. Then you if you make y = 2f(x-6), as following, you obtain:

y = 2f(x-6)=2[x-6] (The function It expands vertically in the y-axis).

4. That is how you obtain the final function.

Answer: 1/4 I think

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

It actually trying to trick you with a double negative

Answer:

A

Step-by-step explanation:

Squares do have right angles, if they did not have to then all other ssquares with no right angles would be considered rhombuses.

The question is sking which does not apply, a does not apply but thats what theyre asking so its a

Answer:

42

Step-by-step explanation:

Let that number be x

x is multiplied by 0.5, 1/3, then squared

(0.5/3 x)^2

Add 1: (0.5/3 x)^2 +1

Make it equal to 50

: (0.5/3 x)^2 +1 = 50 now solve

(0.5/3 x)^2 =50-1

(0.5/3 x)^2 =49

square root both sides ending up with:

0.5/3 x=7

solve for x

x=42

GOOD LUCK AND HOPE IT HELPS:))

Answer:

8cm

Step-by-step explanation:

8×8×8

[tex] {8}^{3} {cm}^{3} [/tex]

Answer:shift two units to the right

Step-by-step explanation:

Answer: x= 1/2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x−5(x−1)=2x−(2x−3)

x−5(x−1)=2x+−1(2x−3)(Distribute the Negative Sign)

x−5(x−1)=2x+−1(2x)+(−1)(−3)

x−5(x−1)=2x+−2x+3

x+(−5)(x)+(−5)(−1)=2x+−2x+3(Distribute)

x+−5x+5=2x+−2x+3

(x+−5x)+(5)=(2x+−2x)+(3)(Combine Like Terms)

−4x+5=3

−4x+5=3

Step 2: Subtract 5 from both sides.

−4x+5−5=3−5

−4x=−2

Step 3: Divide both sides by -4.

−4x /−4 = −2 /−4

x = 1/2

Answer:

Step-by-step explanation:

It is an arithmetic sequence. The nth term = 499 + n

Answer:

13/6

Step-by-step explanation:

Answer:

18

Step-by-step explanation:

18*5=90

18*7=126

answer:

number 4

Step-by-step explanation:

24 ? 8×3=24

Answer:

a. ∀ a, b ∈ Z, if a < 0 ∧ b < 0 ⇒ a + b < 0

b. ∀ a, b ∈ Z, if a > 0 ∧ b > 0 ⇒ a - b < 0 ∨ a - b > 0

c. ∀ a, b ∈ Z, a² + b² ≥ (a + b)

d. ∀ a, b ∈ Z, |ab| = |a||b|

Step-by-step explanation:

a. The sum of two negative integers is negative.

First, we take the quantifier ∀ and write the two integers a and b as a element of the set of integers Z as ∀ a, b ∈ Z. Now, if a and b are negative, we write if a < 0 ∧ b < 0 where the logical operator ∧ represents "and". So, we have the statements ∀ a, b ∈ Z, if a < 0 ∧ b < 0. Then finally, we use the logical operator ⇒ to imply the summation of the two integers as ⇒ a + b. And since we are to show that their sum is less than zero, we write ⇒ a + b < 0.

Combining this statement with the previous expressions, we have

∀ a, b ∈ Z, if a < 0 ∧ b < 0 ⇒ a + b < 0

b. The difference of two positive integers is not necessarily positive.

First, we take the quantifier ∀ and write the two integers a and b as a element of the set of integers Z as ∀ a, b ∈ Z. Now, if there exists a and b that are positive, we write if a > 0 ∧ b > 0 where the logical operator ∧ represents "and". So, we have the statements ∀ a, b ∈ Z, a > 0 ∧ b > 0. Then finally, we use the logical operator ⇒ to imply the difference of the two integers as ⇒ a - b. And since we are to show that their difference is either less than zero or greater than zero, we write ⇒ a - b < 0 ∨ a - b > 0 where ∨ implies the logical operator "or".

Combining this statement with the previous expressions, we have

∀ a, b ∈ Z, if a > 0 ∧ b > 0 ⇒ a - b < 0 ∨ a - b > 0

c. The sum of the squares of two integers is greater than or equal to the square of their sum.

First, we take the quantifier ∀ and write the two integers a and b as a element of the set of integers Z as ∀ a, b ∈ Z. Now, since we are dealing with the sum of the squares of each integer, we write a² + b². And, we are to show that this sum is greater than the sum of the individual integers, we write their sum as a + b, and we show the previous sum to be greater than or equal to this as a² + b² ≥ (a + b).

Combining this statement with the previous expressions, we have

∀ a, b ∈ Z, a² + b² ≥ (a + b)

d. The absolute value of the product of two integers is the product of their absolute values.

First, we take the quantifier ∀ and write the two integers a and b as a element of the set of integers Z as ∀ a, b ∈ Z. Now, since we are to show that absolute value of the product of the integers equals the product of the absolute value of the individual integers, we write the that absolute value of the product of the integers as |ab| and the product of the absolute value of the individual integers as |a||b|.

Since we are to show that this expression and the previous expression are equal, we write |ab| = |a||b|.

Combining this statement with the previous expressions, we have

∀ a, b ∈ Z, |ab| = |a||b|

Answer:

16 packs of spearmint gum

Step-by-step explanation:

Let x = no. of packs of spearmint gum

2x = no. of packs of green apple gum

2x - 6 = no. of packs of cinnamon gum

x + 2x + 2x - 6 = 74

5x - 6 = 74

5x = 80

x = 16 packs of spearmint gum

2x = 2(16) = 32 packs of green apple gum

2x - 6 = 32 - 6 = 26 packs of cinnamon gum

Answer:

php 1500

24 pesos

Step-by-step explanation:

Purchase price = 7500

Amount sold = 9000

The markup will be :

Amount sold - purchase price

9000 - 7500

= php 1500

2.)

Price = 80 pesos

Markup rate % = 30

The Markup = 30% * 80

Markup = 0.3 * 80

Markup = 24 pesos