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Naily [24]
3 years ago
12

△ABC and △JKL are two triangles such that ∠A≅∠J and ∠B≅∠K. Which of the following would be sufficient to prove that the triangle

s are congruent? A ACJL=BCKL B ∠C≅∠L C AB≅JK D BC≅Jk
Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0
Kanchenjunga are eaywkwk flak an
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QUICK! heres the question. select the equations where x=3 is a soultion
Snowcat [4.5K]

Answer: x-3=0 and 6=2x

Step-by-step explanation: to find what X equals for each equation you have to get X alone to get X alone in the first option you have to do +3 to both sides of the equal sign which leaves you with x=3 which you then do for the rest of the options an select all the one left with x=3.

8 0
3 years ago
Which table does not represent a relationship that is directly proportional? Btw i clicked a by accident
Ad libitum [116K]

Answer:

The selected option

Step-by-step explanation:

Coincidentally, you answered correctly.

From the values of 3 and five, we can assume that the function of the table is Y=x+2.

But if we move to the next value pairs, that function is false.

10=6+2 that isn't correct.

If you apply this to all the pairs in the table you will find that there is not a function that represents the values given. Therefore, the first option is correct.

3 0
3 years ago
how many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt
anygoal [31]

50 ounces of 15% saline solution contains 0.15*50 = 7.5 ounces of salt.

Let x be the amount of pure water you add to this solution. Then you end up with (x + 50) ounces of this new solution. The amount of salt stays the same.

You want the new solution to have a concentration of 10%. This means you need

7.5 / (x + 50) = 0.1

==>  7.5 = 0.1(x + 50)

==>  7.5 = 0.1x + 5

==>  2.5 = 0.1x

==>  2.5/0.1 = 0.1x/0.1

==>  x = 25

7 0
4 years ago
Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

(( \tan {}^{2} (x)  + 2 \tan(x)  \sin(x)  +  \sin {}^{2} (x)  - ( \tan {}^{2} (x)  - 2 \tan(x)  +  \sin {}^{2} (x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

5 0
2 years ago
Find a equation for the slope 3/2 and Y intercept -1
Mashcka [7]

Answer:

\large\boxed{y=\dfrac{3}{2}x-1}

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

<em>m</em><em> - slope</em>

<em>b</em> - <em>y-intercept</em>

We have

m=\dfrac{3}{2},\ b=-1

Substitute:

y=\dfrac{3}{2}x+(-1)=\dfrac{3}{2}x-1

5 0
3 years ago
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