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Naya [18.7K]
2 years ago
8

In a science experiment, Shelly has to weigh a compound in grams. The balance in her school measures in milligrams (mg). Shelly

finds that the weight of the compound is 8,450 mg. What is the value in grams (g) that Shelly should enter?
Mathematics
1 answer:
exis [7]2 years ago
7 0

Answer:

8.45 grams

Step-by-step explanation:

In a science experiment, Shelly has to weigh a compound in grams. The balance in her school measures in milligrams (mg). Shelly finds that the weight of the compound is 8,450 mg. What is the value in grams (g) that Shelly should enter?

1 mg = 0.001 g

8450 mg = x

Cross Multiply

= 8450 × 0.001 g

= 8.45 g

Therefore, Shelly should enter 8.45 grams

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Step-by-step explanation:

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Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t
Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

\frac{dy}{dx}=0 (The slope is zero)

--To obtain vertical tangent lines, the condition is:

\frac{dy}{dx}=\frac{1}{0} (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}=2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0

Solving for dy/dx:

\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}

Applying the first conditon (slope is zero)

\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

4 0
3 years ago
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