The cost of 5 cans of dog food is $4.35. At this price, how much do 11 cans of dog food cost?
2 answers:
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A right triangle will be formed with 2.2 km as the hypotenuse, 0.5 degrees as the angle, and an unknown height (see attachment).
Recall the trigonmetric ratios sine, cosine, and tangent that can be used for right triangles.
Sin<span>θ = opposite side/hypotenuse
</span>Cos<span>θ = adjacent side/hypotenuse
</span>tan<span>θ = opposite side / adjacent side
</span>θ (theta) refers to the angle (not the right angle). In this case, θ is the given angle of 0.5°.
In this problem, you will use sine because there is a side opposite to the angle (the unknown side h) and the hypotenuse, and sine relates those two sides.
Let the variable h represent the unknown opposite side, the height.
sin(0.5°) = h / (2.2km)
h = (2.2km)sin(0.5<span>°)
h = 0.019 km
Hope I helped! If you have questions on any of the steps, please comment below.</span>
Recall the trigonmetric ratios sine, cosine, and tangent that can be used for right triangles.
Sin<span>θ = opposite side/hypotenuse
</span>Cos<span>θ = adjacent side/hypotenuse
</span>tan<span>θ = opposite side / adjacent side
</span>θ (theta) refers to the angle (not the right angle). In this case, θ is the given angle of 0.5°.
In this problem, you will use sine because there is a side opposite to the angle (the unknown side h) and the hypotenuse, and sine relates those two sides.
Let the variable h represent the unknown opposite side, the height.
sin(0.5°) = h / (2.2km)
h = (2.2km)sin(0.5<span>°)
h = 0.019 km
Hope I helped! If you have questions on any of the steps, please comment below.</span>
Answer:
Using either method, we obtain:
Step-by-step explanation:
a) By evaluating the integral:
The integral itself can be evaluated by writing the root and exponent of the variable u as:
Then, an antiderivative of this is:
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and
Since this this function is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: