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Irina-Kira [14]
3 years ago
15

The least. Common multiple for 8 19 and 23

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0
<h2><em>Answer:</em></h2>

The least common multiple for 8, 19, 23 would be 3496

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Alicia is solving the equation shown.
kvasek [131]

Answer:

D.

Step-by-step explanation:

-2 is divided on both sides of the equation, which is the multiplication property of equality. you aren't adding anything/ multiplying anything by 0

6 0
3 years ago
What are the restrictions for y <br> X=10/5+y
Naya [18.7K]

Answer:

y\ne-5

Step-by-step explanation:

So you have the equation:

  x=\frac{10}{5+y}. As you may know, you cannot divide by 0, so y has to be restricted in a way that the denominator is never equal to 0. So to find when y makes the denominator 0, simply set the denominator equal to 0, and solve for y. This gives you the equation: 5+y = 0, y=-5. So the y value CANNOT be equal to -5. Because it makes the denominator 0. so y\ne-5 would be the restriction.

7 0
2 years ago
Share £40 in the ratio 1:4 between Tim and Sam
Sauron [17]

Answer:

Tim gets £10 and sam gets £30

Step-by-step explanation:

7 0
3 years ago
Please Help! Are F(x) and G(x) inverse functions across the domain [3,+∞)?
allochka39001 [22]

Answer:

A

Explanation:

For functions to be inverse, it must be true that f( g(x) ) = x and g( f(x) ) = x.

But for F( G(x) ), we have √( G(x) - 3 ) + 8

= √( (x+8)² - 3 - 3 ) + 8

= √( (x+8)² - 6 ) + 8

This -6 part should be cancelled out for functions to work out but we cannot do that, therefore F(x) and G(x) are not inverse.

5 0
3 years ago
Find the coordinates of the midpoint of LB if L(8,5) and B(-6,2)<br><br> 10 points!!!
KATRIN_1 [288]

Answer:

A, (1, 3+1/2)

Step-by-step explanation:

Midpoint formula for reference: m= {(x1 + x2)/2, (y1 + y2)/2}

Plugging in the points we get: m= {(8 - 6)/2, (5 + 2)/2}

Now we simplify using PEMDAS. First step is parentheses.

m= {2/2, 7/2}

Simplifying again (and making 7/2 a mixed number), it becomes

m= {1, 3+1/2}

Hope this helps!

3 0
3 years ago
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