The area of a rectangle is 1,176 square meters. The width of the rectangle is 9 meters. What is the length of the rectangle?

Mathematics

2 answers:

s2008m [1.1K]3 years ago

8 0

The answer for this question is 16

Brums [2.3K]3 years ago

5 0

A = L * W
A = 1176
W = 9

1176 = 9L
1176 / 9 = L
130.67 = L <=== the length is 130.67 meters (and thats rounded)

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The graphs of g(x) = x³- ax² + 6 and h(x) = 2x² + bx + 3 touch when x = 1. Therefore, the tangent to

Olegator [25]

Answer:

$(1,\, 10)$ .

Step-by-step explanation:

Differentiate each function to find an expression for its gradient (slope of the tangent line) with respect to $x$ . Make use of the power rule to find the following:

$g^\prime(x) = 3\, x^2 - 2\, a\, x$ .

$h^\prime(x) = 2\, (2\, x) + b = 4\, x + b$ .

The question states that the graphs of $g(x)$ and $h(x)$ touch at $x = 1$ , such that $g^\prime(1) = h^\prime(1)$ . Therefore:

$3 - 2\, a = 4 + b$ .

On the other hand, since the graph of $g(x)$ and $h(x)$ coincide at $x = 1$ , $g(1) = h(1)$ (otherwise, the two graphs would not even touch at that point.) Therefore:

$1 - a + 6 = 2 + b + 3$ .

Solve this system of two equations for $a$ and $b$ :

$\begin{aligned}& a + b = 2 \\ & 2\, a + b = -1\end{aligned}$ .

Therefore, $a = -3$ whereas $b = 5$ .

Substitute these two values back into the expression for $g(x)$ and $h(x)$ :

$g(x) = x^3 + 3\, x^2 + 6$ .

$h(x) = 2\, x^2 + 5\, x + 3$ .

Evaluate either expression at $x = 1$ to find the $y$ -coordinate of the intersection. For example, $g(1) = 1 + 3 + 6 = 10$ . Similarly, $h(1) = 2 + 5 + 3 = 10$ .