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seropon [69]
3 years ago
12

Anybody help me out??????

Mathematics
2 answers:
GREYUIT [131]3 years ago
7 0
A is the better choice
matrenka [14]3 years ago
6 0

Answer:

choice A

Step-by-step explanation:

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zvonat [6]

now, if we check what is 36 ÷ 8 = 4.5, so -36/8 = -4.5, and -4.5 = -4.5, so they're both equal, neither is greater.

7 0
4 years ago
16 - 4y when y is 4<br> Some plzzzzz helppppp!!!!
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Answer:

0

Step-by-step explanation:

To solve this equation, first find out 4y, which equals 16. Then do the full equation, 16-16=0.

4 0
3 years ago
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A triangle has sides witg lengths of 5x-7, 3x-4, and 2x-6. What is the perimeter of the triangle?
4vir4ik [10]
Perimeter = <span>5x-7+ 3x-4 + 2x-6 = 10x -17</span><span>
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Mathematical Q PRACTICE Use Number Sense A puppy weighs 12 ounces What fractional part of a pound is this A
Lisa [10]
Do 12/16 which equals 0.75
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Hope this helps!
8 0
3 years ago
Derivatives concept:<br> Exercises using the definition of derivatives:<br><br> (Full development)
krok68 [10]

(A) <em>f(x)</em> = 7 is constant, so <em>f(x</em> + <em>h)</em> = 7, too, which makes <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 0. So <em>f'(x)</em> = 0.

(B) <em>f(x)</em> = 5<em>x</em> + 1   ==>   <em>f(x</em> + <em>h)</em> = 5 (<em>x</em> + <em>h</em>) + 1 = 5<em>x</em> + 5<em>h</em> + 1

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 5<em>h</em>

Then

\displaystyle f'(x) = \lim_{h\to0}\frac{5h}h = \lim_{h\to0}5 = 5

(C) <em>f(x)</em> = <em>x</em> ² + 3   ==>   <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 3 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 3

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ²

\implies\displaystyle f'(x) = \lim_{h\to0}\frac{2xh+h^2}h = \lim_{h\to0}(2x+h) = 2x

(D) <em>f(x)</em> = <em>x</em> ² +<em> </em>4<em>x</em> - 1   ==>   <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 4 (<em>x</em> + <em>h</em>) - 1 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 4<em>x</em> + 4<em>h</em> - 1

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ² + 4<em>h</em>

\implies \displaystyle f'(x) = \lim_{h\to0}\frac{2xh+h^2+4h}h = \lim_{h\to0}(2x+h+4) = 2x+4

5 0
3 years ago
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