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4 years ago

14

Any number that is divisible by 2 is also divisible by 8.

2 answers:

Neporo4naja [7]4 years ago

7 0

You don't have to go any further than 2 itself.

All of the even numbers that are not multiples of 8 are counterexamples.

28 is a good one.

loris [4]4 years ago

3 0

28

28/8=3.5

...................

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Can someone please give me the answer to the second part of this question?<br> Thank You!

Elena-2011 [213]

Answer:

$\theta=30\textdegree$

Step-by-step explanation:

$(ii)\ \sqrt{3}\tan\ \theta=2\sin \theta\\\\substitute\ \tan\theta=\frac{\sin\theta}{\cos\theta}\\\\\sqrt{3}\frac{\sin\theta}{\cos\theta}=2\sin\theta\\\\Divide\ both\ sides\ by\ \sin\theta.\\\\\sqrt{3}\frac{\sin\theta}{\cos\theta\times \sin\theta}=\frac{2\sin\theta}{\sin\theta}\\\\\frac{\sqrt{3}}{\cos\theta}=\frac{2}{1}\\\\Apply\ cross\ multiplication\\\\2\cos\theta=\sqrt{3}\\\\divide\ both\ sides\ by\ 2\\\\\frac{2\cos\theta}{2}=\frac{\sqrt{3}}{2}\\\\\cos\theta=\frac{\sqrt{3}}{2}\\\\$

$\cos\theta=\cos\ 30\textdegree\\\\\theta=30\textdegree$

4 0

3 years ago

Plz don't help whats 2+2

Ratling [72]

Answer:

2+2=4

Step-by-step explanation:

lets say you have

oo+oo=oooo

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8 0

4 years ago

Round 13.2 to a whole number

sertanlavr [38]

round to 13 since you round down.

so the answer is 13

3 0

3 years ago

Hey need help....heheheh Oki do u guys know wut 2.19 times 43.7 is......it's for a test -_-

Vikentia [17]

Answer:

95.703

Step-by-step explanation:

If you could subscribe to penny_pg3d that would be great :>

7 0

4 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