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Soloha48 [4]
3 years ago
14

If PQ and RS intersect to form four right angles which statement is true

Mathematics
2 answers:
Rus_ich [418]3 years ago
7 0
Given that PQ and RS intersect to form four right angles, then they form an angle of 90 degrees, this implies that:
if the meet at point O, 
then;
∠POR=∠POS=∠QOS=∠ROQ=90°
Therefore the correct statement is:
PQ is perpendicular to RS
that is to mean:
PQ⊥RS
The answer is C.
Aliun [14]3 years ago
3 0

Answer:

Given: PQ and RS intersect to form four right angles.

Option C is correct.i.e., PQ ⊥ RS

Because Perpendicular means a line intersecting another at right angles and making 4 right angles between them. Here, also PQ intersect with RS at right angles and makes 4 right angle

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Parallel lines e and f are cut by transversal b what is the value of x?
NikAS [45]

Answer: 25

Step-by-step explanation:

Both lines are being cut by the same transversal so to solve for x, make both equations equal to each other

2x+10 = 3x-15

To isolate x, do the inverse operations (for any side chosen)

Do the inverse operations to move -15 and isolate 3x

Add 15 to 10 (10+15=25)

You now have 2x+25=3x

Now, do the same to combine the x values together

25=3x-2x

You now have 25=1x

Divide by 1 on both sides

25=x

4 0
3 years ago
Read 2 more answers
HELP ASAP PLEASE NEED IT RIGHT NOWWWW THANK YOU.!!
Alex73 [517]

Answer:

it should be 71

Step-by-step explanation:

16+5+5+5+5+5+5+5+5+5+5

+5= 71

16 is the 4th and you had 5 eleven more times becuase the numbers are increasing by 5.

5 0
1 year ago
Any one know the answer it has to be turned in tomorrow morning
Veseljchak [2.6K]
17: 900 because 3,000 + 900 + 40 +7 is 3,947 or 3,947-3,000-40-7= 900
6 0
3 years ago
Read 2 more answers
F(x)=x-1/x^2-x-6 which is the graph of
hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

f\left(x\right)=x-\frac{1}{x^2}-x-6

x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:

\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0

-\frac{1}{x^2}-6=0

-1-6x^2=0

\mathrm{No\:Solution\:for}\:x\in \mathbb{R}

\mathrm{No\:x-axis\:interception\:points}

y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:

y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0

As we know that the domain of a function is the set of input or argument values for which the function is real and defined.

\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}

\mathrm{Since}\:x=0\:\mathrm{is\:not\:in\:domain}

\mathrm{No\:y-axis\:interception\:point}

\mathrm{Asymptotes\:of}\:-\frac{1}{x^2}-6:\quad \mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-6

\mathrm{Range\:of\:}-\frac{1}{x^2}-6:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)

The graph is attached below.

5 0
3 years ago
I need help solving these problems.
PolarNik [594]
I can’t see the picture very well
5 0
3 years ago
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