Mathematics, 25.09.2019 00:00 markusblazer
Eight minus the product of two and a number x.
Answers: 1
Mathematics, 21.06.2019 21:30, shelbysargent11
Complete each statement from the information given and the triangle criterion you used. if the triangles cannot be shown to be congruent, leave the box for the second triangle blank and choose for reason “cannot be determined.” carbon - regular hexagon. ∆can ≅ ∆ by
Answers: 1
Mathematics, 21.06.2019 22:30, kaylaamberd
Maria found the least common multiple of 6 and 15. her work is shown below. multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . multiples of 15: 15, 30, 45, 60, . . the least common multiple is 60. what is maria's error?
Answers: 1
Mathematics, 21.06.2019 23:30, jdenty3398
The graph of the the function f(x) is given below. find [tex]\lim_{x \to 0\zero} f(x)[/tex] [tex]\lim_{x \to 1+\oneplus} f(x)[/tex] [tex]\lim_{x \to 0-\zeroneg} f(x)[/tex]
Answers: 1
Mathematics, 22.06.2019 02:00, Zaxx2974
Part a what is the area of triangle i? show your calculation. part b triangles i and ii are congruent (of the same size and shape). what is the total area of triangles i and ii? show your calculation. part c what is the area of rectangle i? show your calculation. part d what is the area of rectangle ii? show your calculation. part e rectangles i and iii have the same size and shape. what is the total area of rectangles i and iii? show your calculation. part f what is the total area of all the rectangles? show your calculation. part g what areas do you need to know to find the surface area of the prism? part h what is the surface area of the prism? show your calculation. part i read this statement: “if you multiply the area of one rectangle in the figure by 3, you’ll get the total area of the rectangles.” is this statement true or false? why? part j read this statement: “if you multiply the area of one triangle in the figure by 2, you’ll get the total area of the triangles.” is this statement true or false? why?
Answers: 1
Eight minus the product of two and a number x....
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