question the base of a triangle is shrinking at a rate of 9 cm/s and the height of the triangle is…

Explanation:

Step1: Recall area formula of triangle

The area formula of a triangle is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height.

Step2: Differentiate with respect to time $t$

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \frac{1}{2}b$ and $v = h$. So $\frac{dA}{dt}=\frac{1}{2}(\frac{db}{dt}h + b\frac{dh}{dt})$.

Step3: Substitute given values

We know that $\frac{db}{dt}=- 9$ cm/s (negative because the base is shrinking), $\frac{dh}{dt}=7$ cm/s, $h = 13$ cm and $b = 4$ cm. Substitute these values into the formula: $\frac{dA}{dt}=\frac{1}{2}((-9)\times13 + 4\times7)$. First, calculate the values inside the parentheses: $(-9)\times13=-117$ and $4\times7 = 28$. Then $(-9)\times13 + 4\times7=-117 + 28=-89$. Finally, $\frac{dA}{dt}=\frac{1}{2}\times(-89)=-\frac{89}{2}=-44.5$ cm²/s.

Answer:

$-44.5$ cm²/s