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3 years ago

What is the domain of the function shown in the table?

Mathematics

1 answer:

Rus_ich [418]3 years ago

5 0

Answer:

C. {3,5,7,9}

Step-by-step explanation:

The domain of a function is the set of inputs, or x-values, of the function.

The x-values listed in this function are 3, 5, 7 and 9; this means the domain is {3, 5, 7, 9}.

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What is the best category for the system of equations shown:

Taya2010 [7]

Answer:

Option (1)

Step-by-step explanation:

System of equations has been given as

2x + 6y = 14

By dividing this equation by 2

x + 3y = 7 --------(1)

Second equation is,

-2x + 2y = -10

-x + y = -5 -------(2)

By adding equation (1) and equation (2)

(x + 3y) + (-x + y) = 7 - 5

2x = 2

x = 1

By putting x = 1 in the equation (1)

1 + 3y = 7

3y = 6

y = 2

So, we get exactly one solution pair (1, 2) where these lines are intersecting.

Therefore, the best category that defines the system of equations is Consistent/Independent

Option (1) will be the answer.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%202%20%20%5Csin%28%20%5Calpha%20%29%20-%20%20%20%5Ccos%28%20%5Calpha%20%29%20%20%3D%202%20%5C

mr Goodwill [35]

$2\sin\alpha-\cos\alpha=2$

Consider the substitution $\tan\dfrac\alpha2=\beta$ . Then by the double angle identities we get

$\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2$

$\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2$

We also have

$\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}$

so that

$\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}$

$\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}$

and the original equation has been transformed to

$\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2$

Solve for $\beta$ :

$5\beta^2-1=2+2\beta^2$

$3\beta^2=3$

$\beta^2=1$

$\beta=\pm1$

Solving for $\alpha$ gives

$\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi$

$\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi$

where $n$ is any integer. Both $\sin$ and $\cos$ are $2\pi$ -periodic, which is to say

$\cos(x+2n\pi)=\cos x$

$\sin(x+2n\pi)=\sin x$

so that

$\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1$

$\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0$

and we find that

$\sin\alpha+2\cos\alpha=\pm1$

3 0

4 years ago

What’s the correct answer for this?

dybincka [34]

Answer:

(A)

Step-by-step explanation:

Using the formula :

Area = 1/2 [-1(1 - 6) -7(6 - 1) -3(1 - 1)]

Area = 1/2 [5 - 35]

Area = 1/2 × -30 = |-15| = 15 units²

5 0

3 years ago

i’m struggling with angles, if anyone wants to message me to help or just explain how to solve this problem, that would be great

shusha [124]

Answer:

what exactly do you need help with?

Step-by-step explanation:

6 0

4 years ago

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Describe the steps for predicting new elements of a sequence.

Brums [2.3K]

Examine the information, verify the conjecture, make a new conjecture

8 0

3 years ago

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