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3 years ago

Solve for x 78.2+0.5x=287

Mathematics

2 answers:

diamong [38]3 years ago

4 0

1. move 78.2 over to the over side. This will give you 287-78.2.

2. when you have .5x = 208.8, divide both sides by .5

3. this should give you x= 417.6

4. make sure to check your work by plugging you answer back into x.

bagirrra123 [75]3 years ago

4 0

x*78.2+0.5=287

x*78.2*10+0.5*10=287*10

782+5=2870

782x+5-5=2865

782x=2865

782x/782=2865/782

2865/782

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Exercise 1: The length of the unknown leg is 4 inches.

Exercise 3: The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4: Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

Step-by-step explanation:

Exercise 1:

Let suppose that triangle represented in the figure is a right triangle, the length of the missing leg is determined by Pythagorean Theorem:

$y = \sqrt{l^{2}-x^{2}}$ (1)

Where:

$l$ - Hypotenuse, in inches.

$x$ - Known leg, in inches.

$y$ - Unknown leg, in inches.

If we know that $l = 11\,in$ and $x = 10\,in$ , then the length of the unknown leg is:

$y = \sqrt{21}$

Since 4 is the least whole number closest to $\sqrt{21}$ , then we conclude that the length of the unknown leg is 4 inches.

Exercise 3:

The range of possible lengths for the missing side of the triangle is represented by the following simultaneous inequality:

$x + y > l > x-y$ (2)

Where:

$x$ - Greater side, in inches.

$y$ - Lesser side, in inches.

$l$ - Missing side, in inches.

If we know that $x = 8\,in$ and $y = 6\,in$ , then we have the following range of missing sides:

$14\,in > l > 2\,in$

The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4:

Let check each pair to determine possible constructions by means of the inequality used in Exercise 3:

(i) $x = 4\,in, y = 2\,in$

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Possible choices: 5 inches.

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$7\,in > l > 3\,in$

Possible choices: 4 inches, 6 inches.

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$8\,in > l > 4\,in$

Possible choices: 5 inches, 6 inches.

(iv) $x = 10\,in, y = 2\,in$

$12\,in > l > 8\,in$

Possible choices: None.

(v) $x = 5\,in, y = 4\,in$

$9\,in > l > 1\,in$

Possible choices: 2 inches, 6 inches.

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$10\,in > l > 2\,in$

Possible choices: 5 inches.

(vii) $x = 10\,in, y = 4\,in$

$14\,in > l > 6\,in$

Possible choices: None.

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$11\,in > l > 1\,in$

Possible choices: 2 inches, 4 inches, 10 inches.

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$15\,in > l > 5\,in$

Possible choices: 6 inches.

(x) $x = 10\,in, y = 6\,in$

$16\,in > l > 4\,in$

Possible choices: 5 inches.

Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

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