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3 years ago

I’ll mark you brainliest!!!!! Which situation shows a constant rate of change?

Mathematics

1 answer:

Irina-Kira [14]3 years ago

4 0

Answer:

Step-by-step explanation:

If we make a graph comparing the total cost of the milk and the gallons bought, it will be a straight line, which is a constant rate of change.

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Verify the trigonometric identity. convert the left side of the relationship to look like the right side.

Tomtit [17]

Answer:

Verified

Step-by-step explanation:

Use $\tan{x}=\frac{\sin{x}}{\cos{x}}$ , $\sin^2{x}+\cos^2{x}=1$ and $\sec{x}=\frac{1}{\cos{x}}$

LHS= $\tan^2{x}+1=\frac{\sin^2{x}}{\cos^2{x}} + 1=\frac{\sin^2{x}+\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}=\sec^2{x}$ =RHS

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3 years ago

Which linear inequality is represented by the graph?

julia-pushkina [17]

Please where is the graph

8 0

3 years ago

A set of data already has the values 11, 14, 23, and 16. What value would have to be added to the set for the mean of the five n

KonstantinChe [14]

1) Take 18 and multiply it by 5 = 90
2) Then subtract 26 = 64
3) Then subtract 18 = 46
4) Then subtract 12 = 34
5) Then subtract 8 = 26

The answer is 26.

5 0

3 years ago

Which of the following methods would be the easiest to use to solve x2 + 5x – 6 = 0?

EastWind [94]

For me personally, the easiest way to do this is by isolating the x² term, and finding the square root of both sides. The hardest way (well actually, the longest way) would be to use the quadratic formula. It just complicates things unnecessarily.

5 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago