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xxTIMURxx [149]
3 years ago
12

Someone help me on this plz

Mathematics
1 answer:
notka56 [123]3 years ago
6 0
The reflection of any point across the x axis is the symmetric of this point, having x axis as axe of symmetry 


Let :

A(1,3) its symmetric vs. x axis is A'(1,-3)

B(-1,5) its symmetric vs. x axis is B'(-1,-5)


C(-3,-3) its symmetric vs. x axis is C'(-3,3)


D(-4,-4) its symmetric vs. x axis is D'(-4,4)

Symmetry vs. the x axis keeps abscises the same & change just the sign of y value 
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The product of 10 and m is less than 22.
Helen [10]
10m≤22 is what it would look like

4 0
4 years ago
How can 3.7 = x + (negative 5) be solved for x in one step?
Nesterboy [21]

Answer:

add five to 3.7

Step-by-step explanation:

aka 3.7+5=x

8.7=x  

i think

3 0
3 years ago
Read 2 more answers
How do you solve this math problem 9x + 3y =3
enot [183]

Answer:

y=1+-2 then simplify

Step-by-step explanation:

9x+3y=3

first dicide so

\frac{9x}{3}  +  \frac{3y}{3}  =  \frac{3}{3}

2x+y=1

now move 2x to other side

so it would turn negative

put in point slope form-y=mx+b

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5 0
4 years ago
Help me out with these 2 questions for 15 points.
jok3333 [9.3K]

Step-by-step explanation:

The time dilation formula is given by

F(t) = \dfrac{t}{\sqrt{1-v^2}}

where t is the time measured by the moving observer and F(t) is the time measured by the stationary earth-bound observer and <em>v</em><em> </em> is the velocity of the moving observer expressed as a fraction of the speed of light.

a) If the observer is moving at 80% of the speed of light and observes an event that lasts for 1 second, a stationary observer will see the same event occurring over a time period of

F(t) = \dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}}

\:\:\:\:\:\:\:=\dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}} =\dfrac{1\:\text{s}}{\sqrt{1-(0.64)}}

\:\:\:\:\:\:\:=1.67\:\text{s}

This means that any event observed by this moving observer will be seen by a stationary observer to occur 67% longer.

b) Given:

t = 1 second

F(t) = 2 seconds

We need to find the speed of the observer such that an event seen by this observer will occur twice as long as seen by a stationary observer. Move the term containing the radical to the left side so the equation becomes

\sqrt{1-v^2} = \dfrac{t}{F(t)}

Take the square of both sides, we get

1 - v^2 =\dfrac{t^2}{F^2(t)}

Solving for <em>v</em>, we get

v^2 = 1 - \dfrac{t^2}{F^2(t)}

or

v = \sqrt{1 - \dfrac{t^2}{F^2(t)}}

Putting in the values for t and F(t) we get

v = \sqrt{1 - \dfrac{(1\:\text{s})^2}{(2\:\text{s})^2}}

v = \sqrt{1 - \dfrac{1}{4}} = \sqrt{0.75}

\:\:\:\:=0.866

This means that the observer must moves at 86.6% of the speed of light.

6 0
3 years ago
What is the image of (2,8) after a reflection over the line y = x?<br> PLS ANSWER FAST
Rina8888 [55]

Answer:

4,16

Step-by-step explanation:

Multiply each by 2.

2 x 2 = 4

8 x 2 = 16

5 0
3 years ago
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