Find the area of the figure. please need help

At one store, Julia saved $10 with a coupon for 40% off her total purchase. At another store, the $29 she spent on books is

barxatty [35]

Let "a" and "b" represent the values of the first and second purchases, respectively.
0.40*(original price of "a") = $10
(original price of "a") = $10/0.40 = $25.00 . . . . divide by 0.40 and evaluate
a = (original price of "a") - $10 . . . . . . Julia paid the price after the discount
a = $25.00 -10.00 = $15.00

At the other store,
$29 = 0.58b
$29/0.58 = b = $50 . . . . . . . divide by the coefficient of b and evaluate

Then Julia's total spending is
a + b = $15.00 +50.00 = $65.00

Julia spent $65 in all at the two stores.

6 0

3 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$