Pracuco shoot #2

Mathematics

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer:

4 teaspoons of salt.

Step-by-step explanation:

2+2=4

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The length of a spring varies directly with the mass of an object that is attached to it. When a 30-gram object is attached, the

Korvikt [17]

Answer:

Step-by-step explanation:

I'm assuming that the length of the spring is s and the mass of the object is m. If that be the case, the direct variation equation, which a line btw, is

s = km where k is the constant of proportionality. We have to solve for k, the slope of the line, in order to determine the model for this situation.

9 = k(30) so

$k=\frac{9}{30}=\frac{3}{10}$ Now that we know, the slope of the line, we can rewrite the equation with the slope in place:

$s=\frac{3}{10}m$ , choice A.

4 0

3 years ago

A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by:

Kryger [21]

Answer:

Yes

8 ft

Step-by-step explanation:

The equation of the arch is

$y=-0.0625x^2+1.25x+5.75$

Differentiating with respect to $x$ we get

$\dfrac{dy}{dx}=-2\times 0.0625x+1.25=-0.125x+1.25$

Equating with zero

$0=-0.125x+1.25\\\Rightarrow x=\dfrac{-1.25}{-0.125}\\\Rightarrow x=10$

Double derivative of the equation

$\dfrac{d^2y}{dx^2}=-0.125$

So, the function is maximum at $x=10$

$y=-0.0625x^2+1.25x+5.75=-0.0625\times 10^2+1.25\times 10+5.75\\\Rightarrow y=12$

The function is maximum at $(10,12)$

So, the truck's center should be below the point $(10,12)$ for maximum width to pass through.

Now truck is 11 feet tall so $y=11$

$11=-0.0625x^2+1.25x+5.75\\\Rightarrow -0.0625x^2+1.25x-5.25=0\\\Rightarrow x=\frac{-1.25\pm \sqrt{1.25^2-4\left(-0.0625\right)\left(-5.25\right)}}{2\left(-0.0625\right)}\\\Rightarrow x=6,14$