A farm stand wants to sell fruit to local grocery stores in packages with

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

A.12 grapes, 23 plums

Step-by-step explanation:

1. solve all the equations by multiplying each number by how much each item is worth

2. A: 290 =(12× 5)+(10×23)

B: 100 =(10×5)+(5×10)

C:125 =(5×5)+(10×10)

D:235 =(23×5)+(12×10)

You might be interested in

Triangle ABC is graphed in Quadrant I of a coordinate plane, and none of the vertices have an x-coordinate or

Ulleksa [173]

Answer:

Step-by-step explanation:

D. the corresponding sides of triangle ABC and triangle A'B'C' are congruent

because the sides do not change their length when we rotate the triangle

8 0

2 years ago

Does anyone know how to answer this exercise?

neonofarm [45]

Answer:

Step-by-step explanation:

The leftmost line segment has a slope of -2 and (-1,-1) is a point on the line.

Equation: y = -2x-3

middle line segment has a slope of 0 and y-intercept of -2.

Equation: y = -2

The rightmost line segment has a slope of 1.5. (5,0) is a point on the line.

y = 1.5(x-5) = 1.5x-7.5

8 0

2 years ago

Select the relations that are functions.

Arlecino [84]

Answer:

Depending on how the input of each function defined,

The first choice $\left\lbrace(a, 1),\, (6, 1), \, (C, 1)\right\rbrace$ ,
The third choice $\left\lbrace(1, a),\, (2, a), \, (3, a)\right\rbrace$
The fourth choice $\left\lbrace(a, a),\, (b, b), \, (c, c)\right\rbrace$

might be functions.

Step-by-step explanation:

A function between two sets (domain and range) should

be defined for all elements in the domain, and
map each element from the domain to exactly one element in the range.

The second choice can't be a function since the element $a$ from the domain is mapped to more than one element in the range.

Keep in mind that a function should be defined for all elements in its domain. For the first relation to be a function, its domain needs to be $\lbrace a,\, 6, \, C\rbrace$ . Similarly, the domain for the third and fourth relations should be $\lbrace 1,\, 2, \, 3\rbrace$ and $\lbrace a,\, b, \, c\rbrace$