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3 years ago

Problem 3 erasers cost $4.41 Which equation would help determine the cost of 4 erasers?

Mathematics

2 answers:

Phantasy [73]3 years ago

5 0

Answer:

4.41/ 3 = 1.47 each

Step-by-step explanation:

il63 [147K]3 years ago

5 0

The answer is 1.47 each erasers
:)

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Generate an equivalent fraction for 3/4.prove answer with visuals

miv72 [106K]

Step-by-step explanation:

3/4 × 3/3 = 9/12

3/4 × 2/2 = 6/8

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3 years ago

Rectangle ABCD is shown to the right. Point E is located

mafiozo [28]

A) The area of the rectangle ABCD is base times height. 20in x 12in = 240 in sq.
B) The area of the triangle AED is base times height divided by two. (14in x 12in)/2 = 84in sq.
C) Figure EBCD is a right trapezoid (having one pair of parallel sides, while the others are slanted and forming a right angle).
D) Area of EBCD is the area of the rectangle minus the area of the triangle. 240in sq - 84in sq = 156in sq.

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2 years ago

Solve for x and y y+2x=2 y=-4x+10

Katena32 [7]

Answer:

x=4 and y=(-6)

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

7 0

3 years ago

Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p

zalisa [80]

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

Statement 1: "The pen is 3 feet wider than it is long"

$Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3$

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Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"

$Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y$

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Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."

$Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\$

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Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."

$Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\$