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3 years ago

Domain and range for y=2+x^2

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

7 0

The domain is the x-axis, and the range is the y values. The domain would be
-∞ to ∞. And the range is 2 to ∞.

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Please help me......

Scilla [17]

Answer:

-1/4

Step-by-step explanation:

I'm not sure what exactly you needed. But this is the slope of the line. Hope this helps! Have a great day! :)

3 0

3 years ago

Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A

liberstina [14]

Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality $\sin ^2\theta +\cos ^2\theta \geq 1$

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

$\sin \theta$ = side opposite to $\theta$ /hypotenuse

$\cos \theta$ = side adjacent to $\theta$ /hypotenuse

$\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta$

$\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta$

So, coordinates of A = $\left ( \cos \theta ,\sin \theta \right )$

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by $x^2+y^2=1$

Put $(x,y)=\left ( \cos \theta ,\sin \theta \right )$

$\cos ^2\theta +\sin ^2\theta =1$

So, $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ satisfies the equation $x^2+y^2=1$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ inside the circle, $\cos ^2\theta +\sin ^2\theta$

For points $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ outside the circle, $\cos ^2\theta +\sin ^2\theta >1$

So, only points on the circle satisfy the given inequality.

4 0

2 years ago

Read 2 more answers

Please tell me what y equals please

bagirrra123 [75]

The data cannot be modeled by a linear equation because the rate of change is not constant. This can be seen in the fact that when you increase x from 1 to 2, you subtract 3 from the y-values. However, as you add 2 more, you continue to subtract 3 from the y-values. If the function were a linear function, we would likely subtract 6 from the y-values, since that would form a constant ratio.

You can even look at the fact that as we increased the x by 4, we continued to decrease the y by 3. This does not form a constant ratio, and is thus the data could not be modeled by a linear function.

5 0

3 years ago

Plz help me on this question

Sladkaya [172]

y is the weight in kg

So sub 63 in y

63=0.91x-65.5

63+65.5=0.91x

128.5=0.91x

128.5÷0.91=x

141(3 s.f.)=x

7 0

3 years ago

Find 2a for a = 3 1/4. 23 1/4 6 1/4 6 1/2 5 1/2

-BARSIC- [3]

Answer:

6 $\frac{1}{2}$

Step-by-step explanation:

you are adding 3 1/4 to 3 1/4 which is 6 2/4 or 6 1/2

or you can multiply 13/4 by 2 to get 26/4 or 13/2 which, again, is 6 1/2

7 0

3 years ago