Ask question

Ask question

All categories

3 years ago

14

if k is a half-plane determined by QR, then for every real number 0 < x < or = to 180, there is exactly one ray, QQP, that

lies in k such that m < PQR = ______

1 answer:

zloy xaker [14]3 years ago

8 0

Let me start from the meaning of half plane.you must have learned that there are two axes X and Y axes which includes positive X axes and negative X axes and there is positive Y axes and negative y axes. So there are four quadrants which i have explained in the figure. I have explained the meaning of Half plane in the Diagram also. Half plane=Figure obtained by covering any two adjacent regions of the coordinate axes.So we get four half planes.

Now your question is k is a half plane determined by QR, such that 0<x≤180 and there is exactly one ray QRP that lies in k

So, $0< \angle PQR \leq 180^\cir{0}$

You might be interested in

Which of these statements is true for f(x)=(1/10)^x

lana66690 [7]

Step-by-step explanation:

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

Analyzing option A)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

Putting $x = 1$ in the function

$f\left(1\right)=\:\left(\frac{1}{10}\right)^1$

$f\left(1\right)=\:\left\frac{1}{10}\right$

So, it is TRUE that when $x = 1$ then the out put will be $f\left(1\right)=\:\left\frac{1}{10}\right$

Therefore, the statement that '' The graph contains $\left(1,\:\frac{1}{10}\right)$ '' is TRUE.

Analyzing option B)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

The range of the function is the set of values of the dependent variable for which a function is defined.

$\mathrm{The\:range\:of\:an\:exponential\:function\:of\:the\:form}\:c\cdot \:n^{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)>k$

$k=0$

$f\left(x\right)>0$

Thus,

$\mathrm{Range\:of\:}\left(\frac{1}{10}\right)^x:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(0,\:\infty \:\right)\end{bmatrix}$

Therefore, the statement that ''The range of $f(x)$ is $y > \frac{1}{10}$ " is FALSE

Analyzing option C)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

The domain of the function is the set of input values which the function is real and defined.

As the function has no undefined points nor domain constraints.

So, the domain is $-\infty \:$

Thus,

$\mathrm{Domain\:of\:}\:\left(\frac{1}{10}\right)^x\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

Therefore, the statement that ''The domain of $f(x)$ is $x>0$ '' is FALSE.

Analyzing option D)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

As the base of the exponential function is less then 1.

i.e. 0 < b < 1

Thus, the function is decreasing

Also check the graph of the function below, which shows that the function is decreasing.

Therefore, the statement '' It is always increasing '' is FALSE.

Keywords: function, exponential function, increasing function, decreasing function, domain, range

Learn more about exponential function from brainly.com/question/13657083

#learnwithBrainly

3 0

4 years ago

Read 2 more answers

If f(x)=2x(With an exponent of 2) -3, find f(-2)

Mrac [35]

Answer:

5

Step-by-step explanation:

Put -2 where x is and do the arithmetic.

f(x) = 2x² -3

f(-2) = 2(-2)² -3 = 2(-2)(-2) -3 = 8 -3

f(-2) = 5

3 0

4 years ago

Read 2 more answers

Simplify. 12x^4y^5+8x^3y^7-16x^2y^6 over 4xy^5

Rina8888 [55]

Answer:

2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A lot of 200 chips contain 5 that are defective. 8 are selected random with out replacement.

vaieri [72.5K]

<u>Answer:</u>

1 , $\frac{1}{66}$ (Answer)

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

Since, 8 chips are selected at random and 5 are there defectives in the lot, So, at least (8 - 5) = 3 chosen chips will be non - defective.

So, P ( At least two of selected part is non defective ) = 1 .

P(first and second samples are defective)

= $\frac{5 \times 4}{200 \times 199}$

P ( first, second and third samples are defective)

= $\frac{5 \times 4 \times 3}{200 \times 199 \times 198}$

So,

$\frac{{\textrm{Third sample is defective | first and second samples are defectives}}}$

= $\frac{{\textrm{ P ( first, second and third samples are defective)}}}{{\textrm{ P(first and second samples are defective) }}}$

= $\frac{3}{198}$

= $\frac{1}{66}$ (Answer)

7 0

4 years ago

The product of 7/8 and 9/10 is greater than 9/10

sweet [91]

Answer:

No

Step-by-step explanation:

7/8x9/10=63/80

So to compare they need to have the same demoninators

9/10x8/8=72/80

8 0

3 years ago