![ddavid9361](/avatars/45506.jpg)
ddavid9361
18.03.2021 •
Mathematics
What is the range of the function graphed below?
Solved
Show answers
More tips
- C Computers and Internet Best Applications for Your iPad: Review of the Best Candidates for Installation...
- S Style and Beauty Intimate Haircut: The Reasons, Popularity, and Risks...
- A Art and Culture When Will Eurovision 2011 Take Place?...
- S Style and Beauty How to Choose the Perfect Hair Straightener?...
- F Family and Home Why Having Pets at Home is Good for Your Health...
- H Health and Medicine How to perform artificial respiration?...
- H Health and Medicine 10 Tips for Avoiding Vitamin Deficiency...
- F Food and Cooking How to Properly Cook Buckwheat?...
- F Food and Cooking How Many Grams Are In a Tablespoon?...
- L Leisure and Entertainment Carving: History and Techniques for Creating Vegetable and Fruit Decorations...
Answers on questions: Mathematics
- M Mathematics Which of the following expressions has a simplified value of 3a – 2? HELP ME PLEASE X(...
- M Mathematics Which graph shows the solution to this system of inequalities? y 3x - 4 y x+1...
- M Mathematics A car has a mass of 19.3 kg and an acceleration of 18.93 m/sec/sec. What is the force on the car?...
- M Mathematics Solve the inequality -25/12 ≤ v + 5/3...
- M Mathematics I need help with this please...
- M Mathematics Free points for helping me pass m exam btw I made 100!...
- M Mathematics raymond opens a car wash and keep track of his weekly earnings, as shown on the table. Use the linear model to make a prediction and classify the prediction as a interpolation or...
- M Mathematics 2x^2−16x−4=0 quadratic equation...
- M Mathematics Pls help due at 11:59...
- M Mathematics Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse...
Ответ:
Step-by-step explanation:
As we see on the graph, the range is:
(-∞, 2) ory < 2Ответ:
12x=900 Divide both sides by 12.
x=75 hammers to begin with.