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rosijanka [135]
3 years ago
6

Evaluate the expression. 2 • 15 – 7 + 4

Mathematics
1 answer:
weqwewe [10]3 years ago
3 0
2*15-7+4
30-7+4
23+4
27
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3n-2=0 and 4n+1=0
solve

3n-2=0
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n=2/3

4n+1=0
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n=-1/4


n=2/3 and/or -1/4
7 0
3 years ago
In Exercise,find the horizontal asymptote of the graph of the function.<br> f(x) = 7/1+5x
S_A_V [24]

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You compare the degrees of the numerator and the denominator.

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6 0
3 years ago
Which fraction is close to, but less than 1/2?
qwelly [4]

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1/9, 2/9, 3/9, 4/9, 1/8, 2/8, 3/8, 1/7, 2/7, 3/7, 1/6, 2/6, 1/5, 2/5, 1/4, 1/3.

Step-by-step explanation:

hope this helps,

8 0
3 years ago
What is the least common multiple of 6, 16, and 24?
Andreas93 [3]
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5 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Oduvanchick [21]

Answer:

  see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

  c² = a² +b² . . . . . legs a, b; hypotenuse c

__

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

  (√c)² = (√a)² +(√b)²

  c = a +b

That is ...

  legs √a, √b ⇒ hypotenuse √(a+b)

__

<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

6 0
1 year ago
Read 2 more answers
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