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2 years ago

The length of a rectangle is five more than triple the width. If the perimeter is 154 inches, what are the dimensions

Mathematics

1 answer:

Mkey [24]2 years ago

8 0

Answer:

Width = 18 inches

Length = 59 inches

Step-by-step explanation:

The length of a rectangle is five more than triple the width. If the perimeter is 154 inches, what are the dimensions

Perimeter of a rectangle = 2L + 2W

The length of a rectangle is five more than triple the width.

L = Length = 5 + 3W

W = Width

If the perimeter is 154 inches, what are the dimensions

Hence,

154 = 2L + 2W

154 = 2(5 + 3W ) + 2W

154 = 10 + 6W + 2W

Collect like terms

154 - 10 = 8W

144 = 8W

W = 144/8

W = 18 inches

Length = 5 + 3W

Length = 5 + 3(18)

Length = 5 + 54

Length = 59 inches

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VashaNatasha [74]

Answer:

According what I can read, I have the following statements:

$\lim_{x \to 2} f(x) = 1$

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c) Applying properties of limits

$\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)} = \sqrt{1} = 1$

d) Applying properties of limits

$\lim_{x \to 2} 4*g(x)*f(x) = 4*\lim_{x \to 2} g(x)*\lim_{x \to 2} f(x) = 4*-4*1 =-16$

e) Applying properties of limits

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3 years ago

11th grade geometry:

Aliun [14]

Answer:

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Step-by-step explanation:

$\triangle SBA$ has coordinates $S(15,-8),B(-2,21)$ and $A(0,0)$

Using the distance formula.........

Length of side $SB = \sqrt{(15+2)^2+(-8-21)^2}= \sqrt{17^2+(-29)^2}= \sqrt{1130}$

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So, the perimeter of the triangle will be: $(SB+BA+AS)= \sqrt{1130}+ \sqrt{445}+17 =71.71... \approx 72$ units. (Rounded to the nearest unit)

The height of the triangle for the corresponding base $SB$ is 8.89 units.

Formula for the Area of triangle, $A= \frac{1}{2}(base\times height)$

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3 years ago

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Alexeev081 [22]

Answer:

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3 years ago