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3 years ago

Tell whether each of the following is a function.

1 answer:

n200080 [17]3 years ago

6 0

Options A, C and D are functions.

In a function, for every number for 'x' there is only one 'y'. So, if you have two ordered pairs in a choice that have the same number for 'x', that is not a function.

Remember an ordered pair is written (x, y).

B is not a function as:
(x, y)
(1, -2)
(1, -3)
(1, -4)
^ All of the x-coordinates are the same.

E is not a function as:
(x, y)
(6, -1)
(6, -8)
^ Two of the x-coordinates are the same.

Thus, A, C and D are functions.

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Find m∠STU and m∠TUA

PIT_PIT [208]

Answers:

angle STU = 65 degrees
angle TUA = 123 degrees

=========================================================

Explanation:

Let's find the expression for angle TUS in terms of x

Angle TUA has the expression 11x+2. It will add to angle TUS to get 180 degrees

(angle TUS) + (angle TUA) = 180

angle TUS = 180 - (angle TUA)

angle TUS = 180 - (11x + 2)

angle TUS = 180 - 11x - 2

angle TUS = -11x + 178

Now focus on the interior angles of triangle TUS. We have these three angles

T = 5x+10
U = -11x+178
S = 58

For any triangle, the interior angles always add to 180, so,

T+U+S = 180

(5x+10) + (-11x+178) + (58) = 180

(5x-11x) + (10+178+58) = 180

-6x + 246 = 180

-6x = 180-246

-6x = -66

x = -66/(-6)

x = 11

Use this x value to find the angle measures we're after:

angle STU = 5x+10 = 5*11+10 = 55+10 = 65 degrees
angle TUA = 11x+2 = 11*11+2 = 121+2 = 123 degrees

--------------------------

As an alternative route, you can apply the remote interior angle theorem. This says that adding two interior angles leads to the exterior angle that isn't adjacent to either one.

In this case, we would say

(angle STU) + (angle TSU) = angle TUA

that leads to the equation

(5x+10) + (58) = 11x+2

Solving this should lead to x = 11 to generate the angles mentioned earlier.

6 0

3 years ago

Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .