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3 years ago

The perimeter of a rectangle is 42 inches. If the width of the rectangle is 6 inches, what is the length?

Mathematics

2 answers:

Andrews [41]3 years ago

4 0

Answer:

B. 15 inches.

Step-by-step explanation:

A rectangle shares two pairs of congruent opposite sides.

Perimeter = 42

Let:

the length of the rectangle = l

the width of the rectangle = w = 6

Perimeter = l + l + w + w

Plug in 42 for perimeter & 6 for w in the equation:

42 = l + l + 6 + 6

Combine like terms:

42 = ( l + l ) + (6 + 6)

42 = 2l + 12

Next, isolate the variable, l. Do the opposite of PEMDAS. First, subtract 12 from both sides:

42 (-12) = 2l + 12 (-12)

42 - 12 = 2l

30 = 2l

Next, divide 2 from both sides:

(30)/2 = (2l)/2

l = 30/2

l = 15

B. 15 inches is the length of the rectangle.

Check:

l + l + w + w = 42

15 + 15 + 6 + 6 = 42

30 + 12 = 42

42 = 42 (True).

Stella [2.4K]3 years ago

3 0

Answer:

b.15 inches

Step-by-step explanation:

subtract 12 (2 6inch sides) from the perimeter

get 30 and divide by 2 to get 2 sides

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All you do is referring to the following definition:

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3 years ago

Write the equation of the line shown in point-slope form. (2,1) (3,8)

Ulleksa [173]

Answer:

The line equation that passes through the given points is 7x – y = 13

Explanation:

Given:

Two points are A(2, 1) and B(3, 8).

To find:

The line equation that passes through the given two points.

Solution:

We know that, general equation of a line passing through two points (x1, y1), (x2, y2) in point slope form is given by

$\frac{(y- y1)}{(x-x_1)}= \frac{((y_2- y_1)}{(x_2- x_1 )}$

${(y- y1)= \frac{((y_2- y_1)}{(x_2- x_1 )}\times(x-x_1)$ ..........(1)

here, in our problem x1 = 3, y1 = 8, x2 = 2 and y2 = 1.

Now substitute the values in (1)

$(y-8) = \frac{(1 - 8)}{(2 - 3)}\times(x- 3)$

$(y -8) = \frac{(- 7)}{(-1)}(x-3)$

y – 8 = 7(x – 3)

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3 years ago

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3 years ago

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Hatshy [7]

Answer:

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Step-by-step explanation:

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2 years ago

Write the equation of the line that passes through (0, 3) and is parallel to the line 3x + 5y = 6.

Pavel [41]

Please, see the offered decision:
1) common equation for lines is y=kx+b. If k₁=k₂ (for line 1 and line 2) ⇒ 'line 1' || 'line 2'.
2) for line 3x+5y=6 k= -3/5. It means (according to item 1) for unknown line k is the same (-3/5).
3) using points (0;3) it is easy to find parameter b (x=0, y=3) via y=kx+b:
3=0*(-3/5)+b ⇔ b=3.
4) finaly (k=-3/5; b=3):
$y=- \frac{3}{5}x+3 \ or \ 3x+5y=15$

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3 years ago