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Ostrovityanka [42]
4 years ago
7

I have a test tomorrow

Mathematics
2 answers:
Nana76 [90]4 years ago
5 0

well i wish you the best of luck buddy

balandron [24]4 years ago
5 0
Good luck ❤️❤️❤️

Have a good day
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The point (1,1) is the image under the translation (x,y)-> (x+3,y-3). What is the preimage of this point?
wel

Answer:

(-2,4)

Step-by-step explanation:

\text{pre-image}->\text{image}

(x,y)->(x+3,y-3)

(x,y)->(1,1)

This implies (x+3,y-3)=(1,1).

This means x+3=1 and y-3=1.

Subtract 3 on both sides for first equation: x=1-3=-2.

Add 3 on both sides for the second equation: y=1+3=4.

So the pre-image (x,y) is (-2,4) if the image is (1,1).

Let's check:

(x,y)->(x+3,y-3)

(-2,4)->(-2+3,4-3)

(-2,4)->(1,1)

So the solution has been checked and verified.

6 0
3 years ago
How many times bigger is 6.4*10^9 than 1.3*10^9
sukhopar [10]

Answer:

at least four times bigger

5 0
3 years ago
HELP MEEE PLZ :( Saeed sells beaded necklaces. Each large necklace sells for 6.50 $ and each small necklace sells for4.90$ . How
insens350 [35]
(6.50 * 7) + (4.90 * 6)
45.50 + 29.40 = 74.90
He will earn $74.90
7 0
3 years ago
Read 2 more answers
|x-3|&lt;2<br> |4x+1|&gt;0<br> |x-1|&lt;5<br> Ayudaaaaaaa pleaseeeee
Dafna11 [192]

Recuerda que

• |<em>x</em>| = <em>x</em> si <em>x</em> ≥ 0

• |<em>x</em>| = -<em>x</em> si <em>x</em> < 0

Necesitas considerar dos casos:

• si <em>x</em> - 3 ≥ 0,

|<em>x</em> - 3| < 1   ⇒  <em>x</em> - 3 < 1   ⇒   <em>x</em> < 4

• si <em>x</em> - 3 < 0,

|<em>x</em> - 3| < 1   ⇒   -(<em>x</em> - 3) = 3 - <em>x</em> < 1   ⇒   -<em>x</em> < -2   ⇒   <em>x</em> > 2

Entonces la solución consta de todos los números reales <em>x</em> tales que <em>x</em> > 2 y <em>x</em> < 4, o simplemente 2 < <em>x</em> < 4.

El método para resolver las otras desigualdades es el mismo.

|4<em>x</em> + 1| > 0   ⇒   4<em>x</em> + 1 > 0   o   -(4<em>x</em> + 1) > 0

…   ⇒   4<em>x</em> + 1 > 0   o   -4<em>x</em> - 1 > 0

…   ⇒   4<em>x</em> > -1   o   -4<em>x</em> > 1

…   ⇒   <em>x</em> > -1/4   o   <em>x</em> < -1/4

⇒   <em>x</em> ≠ -1/4

|<em>x</em> - 1| < 5   ⇒   <em>x</em> - 1 < 5   o   -(<em>x</em> - 1) < 5

…   ⇒   <em>x</em> - 1 < 5   o   -<em>x</em> + 1 < 5

…   ⇒   <em>x</em> < 6   o   -<em>x</em> < 4

…   ⇒   <em>x</em> < 6   o   <em>x</em> > -4

⇒   -4 < <em>x</em> < 6

7 0
3 years ago
What is spherical distribution? Specifically in astronomy. HELP!!!!
Yuliya22 [10]

Locating astronomical objects with reference to equatorial coordinate system on the celestial sphere is called spherical astronomy. It is one of the oldest observational astronomical branch which uses various mathematical concepts specifically, geometry calculation for measurement estimation. It can also be referred as positional astronomy generally envelops entire earth geometry to analysis.

Basically, it includes different parameters of earth's geometry along with its equator in line to celestial objects which finally gives location of these objects in particular year. All in all, it is a specialized science field in astronomy which first observes around space envelope and finally gives data about location of individual space objects with respect to tome.

6 0
4 years ago
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