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Elena L [17]
3 years ago
10

A car is traveling at a rate of

Mathematics
2 answers:
Tomtit [17]3 years ago
8 0
63/60 = 1.05 km/min
1.05 * 20 = 21 km/20min
Oxana [17]3 years ago
3 0
63/60min = 1.05k per min
(1.05k/min) * (20min) = 21k
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Which of the following has the greatest value <br>A)3³<br>B)5²<br>C)1¹⁰<br>D)2⁵​
Alecsey [184]

Answer:

A - Highest is 27

Step-by-step explanation:

***If you found this answer helpful please click the heart Icon and rate me 5 star, it would be nice if I got brainliest, thanks!***

7 0
3 years ago
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Determine if the equation below is a function with independent variable x. If so, find the domain. If not, find a value of x to
Andre45 [30]

Answer:

No because it contains points (3,4) and (3,-4).  You cannot have a x assigned to more than one y-value if you want it to be a function.

Step-by-step explanation:

A function has one output per input.

If we are trying to determine if the given is a function of x, then x is the input.

However I can get two outputs from plugging in x=3.

3^2+y^2=25

9+y^2=25

Subtract 9 on both sides:

y^2=25-9

y^2=16

Take the square root of both sides:

y=\pm \sqrt{16}

y=\pm 4.

So input x=3 yields y=4 and y=-4.

Since this input has more than one output then the given is not a function of x.

----Also!

If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:

(\pm 5)^2+y^2=25

25+y^2=25

Subtract 25 on both sides:

y^2=25-25

Simplify:

y^2=0

There is only one value y such that when you square it gives you 0. That is 0.

x=5 only gives y=0 and x=-5 only gives y=0.

There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.

8 0
3 years ago
Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
3 years ago
You found a cute bear so have some free po.ints
Alla [95]

yeah

I understand that I can do you want to go

8 0
2 years ago
Please help!!! I'll mark you brainiest and give you as many Points as possible!!!!!
Vlada [557]

Answer:

Option C

Step-by-step explanation:

f(x) = 40( \frac{1}{4} )^{x}

5 0
3 years ago
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