What is an opposite number?

Answer:

(-3,-2)

(-3,4)

(3,4)

(3,-3)

Step-by-step explanation:

Answer:

cant see nun mind showing it

Answer:

y - 2 = 1/3(x - 3)

Step-by-step explanation:

Point slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Plug in the slope:

y - y1 = m(x - x1)

y - y1 = 1/3(x - x1)

Plug in the given point:

y - y1 = 1/3(x - x1)

y - 2 = 1/3(x - 3)

So, the correct answer is y - 2 = 1/3(x - 3)

Answer:

24

Step-by-step explanation:

4 * 3 * 2 * 1 = 24

Answer:

It's C

Step-by-step explanation:

because there is a 90 degrees sigh behind it and those two angles need to add up to 90

Answer:

31

Step-by-step explanation:

b+59=90

b=90-59

b=31

hope it helps

Answer:

1

Step-by-step explanation:

The average rate of change is the gradient _

Gradient = Rise / Run = (y2 - y1) / (x2 - x1)

From the table ;

x2 = 4 ; x1 = - 1 ; y2 = 5 ; y1 = 0

Gradient = (y2 - y1) / (x2 - x1) = (5-0) / (4 - - 1) = 5/5 = 1

Step-by-step explanation:

we can simplify the stuff inside the parentheses to

[tex] \frac{ {x}^{3} }{2y} [/tex]

now we need to multiply it with itself, giving us

[tex] \frac{ {x}^{6} }{4 {y}^{2} } [/tex]

so yeah, D is the correct answer

Answer:

Step-by-step explanation:

Given that:

[tex]y^2 (y^2-4) = x^2(x^2 -5)[/tex]

at point (0, -2)

[tex]\implies y^4 -4y^2 = x^4 -5x^2[/tex]

Taking the differential from the equation above with respect to x;

[tex]4y^3 \dfrac{dy}{dx}-8y \dfrac{dy}{dx}= 4x^3 -10x[/tex]

Collect like terms

[tex](4y^3 -8y)\dfrac{dy}{dx}= 4x^3 -10x[/tex]

[tex]\dfrac{dy}{dx}= \dfrac{4x^3 -10x}{4y^3-8y}[/tex]

Hence, the slope of the tangent line m can be said to be:

[tex]\dfrac{dy}{dx}= \dfrac{4x^3 -10x}{4y^3-8y}[/tex]

At point (0,-2)

[tex]\dfrac{dy}{dx}= \dfrac{4(0)^3 -10(0)}{4(-2)^3-8-(2)}[/tex]

[tex]\dfrac{dy}{dx}= \dfrac{0 -0}{4(-8)+16}[/tex]

[tex]\dfrac{dy}{dx}= 0[/tex]

m = 0

So, we now have the equation of the tangent line with slope m = 0 moving through the point (x, y) = (0, -2) to be:

(y - y₁ = m(x - x₁))

y + 2 = 0(x - 0)

y + 2 = 0

y = -2

Answer:

consist of either a single capital letter or a capital letter and a small letter.

Step-by-step explanation:

A chemical reaction can be defined as a reaction in which two or more atoms of a chemical element react to form a chemical compound.

In Chemistry, a chemical element can be defined as any pure substance that is typically made up of only atoms and cannot be broken down into simpler substances through chemical processes. Thus, the atomic number (the number of protons in the nuclei of an atom) of a particular chemical element distinguishes from other chemical elements.

Generally, all chemical elements are denoted or represented by a symbol, which may either be single capital letter or a capital letter and a small letter.

This ultimately implies that, the symbols for elements with accepted names consist of either a single capital letter or a capital letter and a small letter. For example, the symbol for sodium is Na, copper is Cu, carbon is C, oxygen is O, iron is Fe, nitrogen is N, magnesium is Mg, potassium is K, argon is Ag, hydrogen is H, helium is He, phosphorus is P, etc.

Answer:

[tex]\displaystyle x \approx 9.9[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Trigonometry

Step-by-step explanation:

Step 1: Define

Identify variables

Angle θ = 64°

Opposite Leg = x

Hypotenuse = 11

Step 2: Solve for x

Answer:

Hi, there the answer is

These are the equations with exactly one solution

-5x + 12 = –12x – 12

-5x + 12 = 5x + 12

-5x + 12 = 5x – 5

Hope This Helps :)

Step-by-step explanation:

First degree equations

A first degree equation has the form

ax + b = 0

There are some special cases where the equation can have one, infinitely many or no solution

If , the equation has exactly one solution

If a=0 and b=0 the equation has infinitely many solutions, because it doesn't matter the value of x, it will always be true that 0=0

If a=0 and  the equation has no solution, because it will be equivalent to b=0 and we are saying it's not true. No matter what x is, it's a false statement.

We have been given some equations, we only need to put them in standard form

-5x + 12 = –12x – 12

Rearranging

7x + 24 = 0

It has exactly one solution because a is not zero

.......................

-5x + 12 = 5x + 12

Rearranging

-10x + 0 = 0

It has exactly one solution because a is not zero

.......................

-5x + 12 = 5x – 5

Rearranging

-10x + 17 = 0

It has exactly one solution because a is not zero

.......................

-5x + 12 = -5x – 12

Rearranging

0x + 24 = 0

It has no solution, no matter what the value of x is, it's impossible that 24=0

Answer: These are the equations with exactly one solution

-5x + 12 = –12x – 12

-5x + 12 = 5x + 12

-5x + 12 = 5x – 5

Answer:

Infinient amount buddy. They are infinent amounts of polygons, as a polygon is a shape with more then 3 sides and closed. This means as long as a number is bigger then 3, there will be a shape for it, so infinent sides

Step-by-step explanation:

Answer:

11.7

Step-by-step explanation:

39.1 - 9.2 = 29.9

29.9 - 18.2 = 11.7

Step-by-step explanation:

Given: [∀x(L(x) → A(x))] →

[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

To prove, we shall follow a proof by contradiction. We shall include the negation of the conclusion for

arguments. Since with just premise, deriving the conclusion is not possible, we have chosen this proof

technique.

Consider ∀x(L(x) → A(x)) ∧ ¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

We need to show that the above expression is unsatisfiable (False).

¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

∃x¬((L(x) ∧ ∃y(L(y) ∧ H(x, y))) → ∃y(A(y) ∧ H(x, y)))

∃x((L(x) ∧ ∃y(L(y) ∧ H(x, y))) ∧ ¬(∃y(A(y) ∧ H(x, y))))

E.I with respect to x,

(L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ ¬(∃y(A(y) ∧ H(a, y))), for some a

(L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ (∀y(¬A(y) ∧ ¬H(a, y)))

E.I with respect to y,

(L(a) ∧ (L(b) ∧ H(a, b))) ∧ (∀y(¬A(y) ∧ ¬H(a, y))), for some b

U.I with respect to y,

(L(a) ∧ (L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

Since P ∧ Q is P, drop L(a) from the above expression.

(L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

Apply distribution

(L(b) ∧ H(a, b) ∧ ¬A(b)) ∨ (L(b) ∧ H(a, b) ∧ ¬H(a, b))

Note: P ∧ ¬P is false. P ∧ f alse is P. Therefore, the above expression is simplified to

(L(b) ∧ H(a, b) ∧ ¬A(b))

U.I of ∀x(L(x) → A(x)) gives L(b) → A(b). The contrapositive of this is ¬A(b) → ¬L(b). Replace

¬A(b) in the above expression with ¬L(b). Thus, we get,

(L(b) ∧ H(a, b) ∧ ¬L(b)), this is again false.

This shows that our assumption that the conclusion is false is wrong. Therefore, the conclusion follows

from the premise.

15

Answer:

30

Step-by-step explanation:

There are 14 data points on the plot

The median is the middle value, which is between the 7th and 8th data points

Since it is even, we find the average of the 7th and 8th data points

Add the 7th and 8Th data points values and divide by 2

( 30+30)/2 = 60/2 = 30

The median is 30

Answer:

30

Step-by-step explanation:

Middle numbers are 30 and 30 so you add them and then divide by two:

[tex]\frac{30 + 30}{2}[/tex] = [tex]\frac{60}{2}[/tex]= 30

Answer:

its c

Step-by-step explanation: