help!!!!!!!! im crying over this stuid question what is the answer

Answer:

x = -1

Step-by-step explanation:

x-7/4=-2

x-7=-8

x=-1

Answer:

There will be 14 and 1/4 GROUPS of 4 toys created by 57 toys

Step-by-step explanation:

57 toys divided into groups of 4 toys.

57/4= 14.25 or 14 1/4 GROUPs

Answer:

3.2

Step-by-step explanation:

3.2/1

Answer:

Step-by-step explanation:

There are 540 degrees in a pentagon so take 540-(everything there)= 67

Answer:

-p^2-6r^2-2r-8

Step-by-step explanation:

When you simplify it you get that. Hope this helped :)

The solution to the given expression is -8 - 2r - p² - 6r² which is determined by performing the arithmetic operation.

Algebraic expressions are mathematical statements with a minimum of two terms containing variables or numbers.

Arithmetic operations can also be specified by subtracting, dividing, and multiplying built-in functions.

The given expression is given in the question

(10 – 7r - p²) + (-6r² - 18 +5r)

Open the parentheses and rearrange the terms,

10 - 18 - 7r + 5r - p² - 6r²

Apply the arithmetic operation in the likewise terms,

-8 - 2r - p² - 6r²

Therefore, the solution to the given expression is -8 - 2r - p² - 6r²

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Answer:

Step-by-step explanation:

Base 5 uses 0, 1 , 2, 3, 4

When we add 2+ 1 = 3

When we add 4 +3 = 7 = 5 +2 . 5 will be carried as 1

4 + 3= 12

In subtraction, 32 - 4

From 2 we cannot subtract 4. So, borrow 5 from 3. So, 2+5 = 7.

7-4=3

32 -4 = 23

I have attached the sums

Answer:

Step-by-step explanation:

g(x^2 + 2x - 4)

3(x^2 + 2x - 4) + 1

3x^2 + 6x - 12 + 1

3x^2 + 6x - 11

Answer:

x = -11

Step-by-step explanation:

-12 = x - 1

-12 + 1 = x

x = -11

Answer:

[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{x = - 11}}}}}[/tex]

Step-by-step explanation:

[tex] \bigstar{ \sf{ \: - 12 = x - 1}}[/tex]

[tex] \tt{Step \: 1 \: }[/tex] : Swap the sides of the equation

[tex] \hookrightarrow{ \sf{x - 1 = - 12}}[/tex]

[tex] \tt{Step \: 2 \: }[/tex] : Move 1 to right hand side and change it's sign

[tex] \hookrightarrow{ \sf{x = - 12 + 1}}[/tex]

[tex] \tt{Step \: 3}[/tex] : Calculate

Remember : The negative and positive integers are always subtracted but posses the sign of the bigger integer.

[tex] \hookrightarrow{ \sf{x = - 11}}[/tex]

The value of x = 11

Hope I helped!

Best regards! :D

~TheAnimeGirl

Answer:

C. -∞ < x < ∞

Step-by-step explanation:

Domain is the set of x-values that can be inputted into function f(x).

We see that our x-values can be any number. Therefore, we have all real numbers as our domain:

(-∞, ∞) or -∞ < x < ∞

Answer:

C. -∞ < x < ∞

Step-by-step explanation:

Plato

g(x) = -(x + 1)^2 - 3

Let's simplify this equation

g(x) = -(x^2 + 2x + 1) - 3

Distribute the negative sign

g(x) = -x^2 - 2x + 1 - 3

Combine like terms

g(x) = -x^2 - 2x - 2

The value of a = -2 and this value is LESS THAN 0

First, let's choose two points from the table to solve for our slope.

(0, -1) and (1, 4)

Slope = (4 -(-1)) / (1 - 0) = 5/1 = 5

The slope of our line is equal to 5

So far we have y = 5x + b

To solve for be, we need to plug one of our points into the equation.

-1 = 5(0) + b

Simplify

-1 = b

So, our equation: y = 5x -1

Answer:

three times a number minus two

Step-by-step explanation:

x= times a number

Answer:

= 37

Step-by-step explanation:

(8+4+3) + (6+5+11)

= 15 + 22

= 37

Answer:C) 4.5 units

Step-by-step explanation:

Answer:

Step-by-step explanation:

The "Commutative Laws" say that when we add then we can swap numbers over and still get the same answer.

Answer:

OP = 4

Step-by-step explanation:

N is the midpoint of MO.

Since NO = 3, then MN = 3, and MO = 3 + 3 = 6

MO + OP = MP

6 + OP = 10

OP = 4

A line segment is also a line but the line with finite length and having fixed endpoints. The length of the line segment OP is 4 units.

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely.

Given the following details as:

Now, from the figure, We have,

Since O is between M and P and N is the midpoint between MO

MP = MN + NO + OP _____(1)

Now,

N is the midpoint of MO so,

MN = N0

N0 = 3 and MP = 10 ____(2)

MN = NO = 3 _______(3)

From (1)

MP = MN + NO + OP

Substituting (2) and (3) in (1)

10 = 3 + 3 + OP

10 = 6 + OP

Subtracting 6 on both sides

10 - 6 = 6 + OP - 6

4 = OP

OP = 4

Thus, OP = 4.

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Marlene needs 15 paper cups to fill with water

An equation is an expression that is used to show the relationship between two or more variables and numbers.

Since each paper cups can hold Two-thirds of a cup of water, hence to hold 10 cups of water:

Number of water cups = 10 cups / (two- thirds) = 10 * 3 / 2 = 15

Marlene needs 15 paper cups to fill with water.

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Answer: The answer is B

Step-by-step explanation: got it right on e2020

Answer: -2

Step-by-step explanation:

Function is missing.

It is;

f(n) = ½⋅f(n − 1) for n ≥ 2

Answer:

Domain for this function is;

n ≥ 1

Step-by-step explanation:

We are told They decide they will have 64 pieces in week 1,

Thus;

f(1) = 64

From the function given;

f(n) = ½⋅f(n − 1)

So;

f(2) = ½•f(2 - 1)

f(2) = ½•f(1)

f(2) = ½ • 64

f(2) = 32

Also;

f(3) = ½•f(3 - 1)

f(3) = ½•f(2)

f(3) = ½ × 32 = 16

Looking at the steps, they follow a general pattern of;

f(n) = 64•½^(n - 1) for n ≥ 1

Thus, the domain for the function for n weeks after the first one is n ≥ 1

Answer:

you subtract the exponents

Answer:

-4x3=12

12÷-6=-2

hope this helps

Answer:

its 28

Step-by-step explanation:

hope this helps also u have to divide 70 divided by 48% in order to get 28

Answer:

18m+n4

Step-by-step explanation:

i think.

Answer: girly im not gonna lie im no mathematician but i came up with 18m  8n i sowy if it's wrong but, you can't blame a girl from trying .

Step-by-step explanation: i combined like terms but im probably still wrong just saying .

Answer:

[tex]x=-1/2=-0.5[/tex]

Step-by-step explanation:

So we have the function:

[tex]h(x)=4x+2[/tex]

To find the zero(s) of a function, set the function equal to 0 and solve for x. So:

[tex]0=4x+2[/tex]

Subtract 2 from both sides:

[tex]-2=4x[/tex]

Divide both sides by 4:

[tex]x=-1/2=-0.5[/tex]

So, the zero of the function is -0.5.

This means that the graph (a line) crosses the x-axis at (-0.5, 0).

Answer:

The answer is

Step-by-step explanation:

3b² - b

when b = 5

Substitute the value of b that's 5 into the expression

That's

3(5)² - 5

= 3(25) - 5

= 75 - 5

We have the final answer as

Hope this helps you

Step-by-step explanation:

-4.5x -2y = -12.5 -------(1)

3.25x -y = -0.75 --------(2)

From (2) we can make y the subject of the relation

y = 0.75+3.25x -------(3)

Input 0.75+3.25x into y in (1)

-4.5x -2(0.75+3.25x) = -12.5

-4.5x -1.5- 6.5x = -12.5

-4.5x -6.5x -1.5 = -12.5

-11x = -12.5 +1.5

-11x = -11

x = 11/11

x = 1

If x = 1,substitute 1 into x in (3)

y = 0.75+3.25x

y = 0.75 +3.25(1)

y = 0.75 +3.25

y = 4

x = 1,y = 4

Answer:

Step-by-step explanation:

the scientific notation of 91,007,500

9.10075 x 10^7

Answer:  see proof below

Step-by-step explanation:

Given: A + B = C                → A = C - B

                                          → B = C - A

Use the Double Angle Identity:     cos 2A = 2 cos² A - 1

                                             → (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS:                     cos² A + cos² B + cos² C

[tex]\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C[/tex]

[tex]\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C[/tex]

[tex]\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C[/tex]

[tex]\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C][/tex]

[tex]\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)[/tex]

[tex]\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)[/tex]

[tex]\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C[/tex]

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   [tex]\checkmark[/tex]