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e-lub [12.9K]
3 years ago
13

Solve: -19 = X - 19 halppppp

Mathematics
2 answers:
Makovka662 [10]3 years ago
7 0

Answer:

0

Step-by-step explanation:

Juliette [100K]3 years ago
3 0

Answer:

x = 0

Step-by-step explanation:

Step 1: Write equation

-19 = x - 19

Step 2: Solve for <em>x</em>

  1. Add 19 to both sides: 0 = x
  2. Rewrite: x = 0

Step 3: Check

<em>Plug in x to verify it's a solution.</em>

-19 = 0 - 19

-19 = -19

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Find the distance between points (5,1) and (3,4) to the nearest tenth
Charra [1.4K]

Answer:

3.6 units (nearest tenth)

Step-by-step explanation:

distance formula

= \sqrt{(x1 - x2)^{2}  + (y1 - y2)^{2} }

Thus, distance between the 2 points

= √[(5 - 3)² + (1 - 4)²]

= √[ 2²+ ( - 3)²]

= √13

= 3.6 units (nearest tenth)

5 0
3 years ago
How are the slopes of parallel lines and perpendicular lines related?
ELEN [110]

Answer:

Let's see what you can do with parallel and perpendicular. In other words, the slopes of parallel lines are equal. Note that two lines are parallel if their slopes are equal and they have different y-intercepts. In other words, perpendicular slopes are negative reciprocals of each other

8 0
3 years ago
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Simplify: StartRoot 324 EndRoot 1. Write the prime factorization of the radicand. StartRoot 324 EndRoot = StartRoot 2 times 2 ti
Savatey [412]

Answer:

  18

Step-by-step explanation:

To continue where you left off, ...

  \sqrt{2^2}\times\sqrt{3^2}\times\sqrt{3^2}=2\times 3\times 3=18

The square root of 324 is 18.

3 0
3 years ago
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PLEASE MATH HELP WILL GIVE BRAINLIEST!
ruslelena [56]
1. y=-3/4x+23/2
2. y=-x/3+10/3
8 0
3 years ago
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Please help i dont understand what its asking
Arisa [49]

Multiply both the numerator and denominator by 2+\sqrt3, which is called the "conjugate" of 2-\sqrt3:

\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}

Why do we pick this number? Recall the difference of squares factorization:

a^2-b^2=(a-b)(a+b)

Now replace a=2 and b=\sqrt3. Then

(2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1

So in your fraction, you end up with

\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)

Finally, just expand the product.

(5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}

7 0
3 years ago
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