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Oduvanchick [21]

3 years ago

10

The length of a rectangle is 4 units less than the width. The area of the rectangle is 21 units. What is the length, in units, o

f the rectangle?

1 answer:

bazaltina [42]3 years ago

4 0

The formula of an area of a rectangle:

A = wl

We have l = w - 4 and A = 21.

Substitute:

w(w - 4) = 21 <em>use distributive property</em>

(w)(w) + (w)(-4) = 21

w² - 4w = 21 <em>subtract 21 from both sides</em>

w² - 4w - 21 = 0

w² - 7w + 3w - 21 = 0

w(w - 7) + 3(w - 7) = 0

(w - 7)(w + 3) = 0 ↔ w - 7 = 0 ∨ w + 3 = 0

w = 7 ∨ w = -3 < 0

l = w - 4 → l = 7 - 4 = 3

<h3>Answer: the length = 3 u.</h3>

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the length of a rectangle is two feet greater then twice its width if the primeter is 25 find the width

Mademuasel [1]

Answer:

Step-by-step explanation:

Perimeter is simply length + width. They gave us that the length is 2 feet plus 2 times the width:

$25=(2w+2)+W$

Subtract 2 from both sides to get 23 = 2w + w

Combine like terms to get 23 = 3w

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3 years ago

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2 years ago

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dezoksy [38]

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3 years ago

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BARSIC [14]

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3 years ago

ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2). Find the coordinates of A' and R

shutvik [7]

Answer:

$A'= (2,-1)$ and $R'=(4,-2)$ under this translation.

Step-by-step explanation:

A translation in $R^{2}$ is a mapping T from $R^{2}$ to $R^{2}$ defined by $T(x,y) = (x + v_1,y+v_2)$ , where $v=(v_1,v_2)$ is a fixed vector in $R^{2}$ .

From the problem we know that $T(2,4)=(3,2)$ , so we need to find the values $v_1$ and $v_2$ such that $T(2,4) = (2 + v_1,4+v_2)=(3,2)$ , so $3=2 + v_1$ and $4+v_2=2$ , thus $v_1=1$ and $v_2=2$ .

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7 0

3 years ago