1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Irina-Kira [14]
2 years ago
5

Please help me please help please help me:) only number 2 I need help

Mathematics
1 answer:
loris [4]2 years ago
6 0

I get you can’t get this I’m in 8th grade it’s hard as infierno what grade are you in

You might be interested in
Need help ASAP !!!!<br> Solve for the missing side. Round answer to two decimal places (like 7.93).
Thepotemich [5.8K]
10.44yd
a^2 + b^2 = c^2
3^2 + 10^2 = c^2
109 = c^2
C (Hypotenuse) ≈ 10.44
3 0
3 years ago
How many solutions does the equation −4y + 4y + 2 = 2 have?
antiseptic1488 [7]

Answer:

D

Step-by-step explanation:

Let's solve your equation step-by-step.

−4y+4y+2=2

Step 1: Simplify both sides of the equation.

−4y+4y+2=2

(−4y+4y)+(2)=2(Combine Like Terms)

2=2

2=2

Step 2: Subtract 2 from both sides.

2−2=2−2

0=0

Answer:

All real numbers are solutions.

8 0
3 years ago
A. f(x) = l-2xl - 3<br> Domain: {1, 2, 3}<br> Find the range
gizmo_the_mogwai [7]

Hello there!

We are given the function:

\displaystyle \large{ f(x) =  | - 2x|  - 3} \\

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

<u>x</u><u> </u><u>=</u><u> </u><u>1</u>

\displaystyle \large{ f(1) =  | - 2(1)|  - 3} \\   \displaystyle \large{ f(1) =  | - 2|  - 3} \\

Recall that any numbers in absolute value are always positive.

\displaystyle \large{ f(1) =  2 - 3} \\   \displaystyle \large{ f(1) =   - 1} \\

<u>x</u><u> </u><u>=</u><u> </u><u>2</u>

\displaystyle \large{ f(2) =  | - 2(2)|  - 3} \\   \displaystyle \large{ f(2) =  | - 4 |   - 3} \\   \displaystyle \large{ f(2) =  4 - 3} \\   \displaystyle \large{ f(2) =  1}

<u>x</u><u> </u><u>=</u><u> </u><u>3</u>

\displaystyle \large{ f(3) =  | - 2(3)|  - 3} \\   \displaystyle \large{ f(3) =  | - 6|   - 3} \\   \displaystyle \large{ f(3) =  6 - 3} \\   \displaystyle \large{ f(3) =  3}

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0
2 years ago
Please help!
nasty-shy [4]
Right change flip it's simple
4 0
3 years ago
If Samantha can pay off her loan in 36 months at a 10% interest rate rather than in 48 months at a 12% interest rate, how much m
Ymorist [56]
Hi there
The simple interest formula is
I=prt
I interest changes
P amount of the loan 6000
R interest rate
T time( number of months/12 months)

The interest in 36 months at a 10%
I=6,000×0.1×(36÷12)=1,800
The interest in 48 months at a 12%
6,000×0.12×(48÷12)=2,880
she will save
2,880−1,800=1,080

Good luck!
7 0
3 years ago
Other questions:
  • Please help. Please. I will mark brainliest
    6·1 answer
  • How many meters equal 9/10 kilometers
    8·2 answers
  • What is the inverse of "if water is ice,then the water's temperature is 32°F
    11·1 answer
  • Find the value of x.
    14·1 answer
  • Jimmy invests $15,000 in an account that pays 8.90% compounded quarterly. How long (in years and months) will it take for his in
    7·1 answer
  • Evaluate the function f(x) = 3x^-2x for x=4
    6·1 answer
  • Which statement is correct? StartFraction 1 inch Over 2.54 centimeters EndFraction = StartFraction 7 inches Over 17.68 centimete
    11·2 answers
  • Linda and Rob open an online savings account that has a 2.89% annual interest rate, compounded
    7·1 answer
  • Find sin(a) <br> (Picture)
    11·1 answer
  • If <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5B%5Bx%2B2%5D%5D" id="TexFormula1" title="f(x)=[[x+2]]" alt="f(x)=[[x+2]]" alig
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!