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3 years ago

2x - 3>5 what the answer

Mathematics

2 answers:

Zepler [3.9K]3 years ago

6 0

The answer is x>0.4
Just like what the other person said

Elden [556K]3 years ago

5 0

So first what you do is add 2x - 3 > 5
Why because do you remember PEMDAS well you have to do it so you have to do it the other way I mean SADMEP. You start with subtraction and you got 2
now what you got is 2x > 5 Now you do the opposite of multiple so you divide 2 divided by 5 is equal: x > 0.4

Hope I had helped

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Given that lim x → 2 f ( x ) = 1 lim x → 2 g ( x ) = − 4 lim x → 2 h ( x ) = 0 limx→2f(x)=1 limx→2g(x)=-4 limx→2h(x)=0, find the

VashaNatasha [74]

Answer:

According what I can read, I have the following statements:

$\lim_{x \to 2} f(x) = 1$

$\lim_{x \to 2} g(x) = -4$

$\lim_{x \to 2} h(x) = 0$

a) Applying properties of limits

$\lim_{x \to 2} f(x) + 5g(x) = \lim_{x \to 2} f(x) + 5 \lim_{x \to 2} g(x) = 1 + 5*-4 = -19$

b) Applying properties of limits

$\lim_{x \to 2} g(x)^{3} = {(\lim_{x \to 2} g(x))}^{3} = (-4)^{3} = -64$

c) Applying properties of limits

$\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)} = \sqrt{1} = 1$

d) Applying properties of limits

$\lim_{x \to 2} 4*g(x)*f(x) = 4*\lim_{x \to 2} g(x)*\lim_{x \to 2} f(x) = 4*-4*1 =-16$

e) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x) = -4*0 =0$

f) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x)*f(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x)*\lim_{x \to 2} f(x = -4*0*1 =0$

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3 years ago

Which ratio is equivalent to the ratio 2:5? a. 6:15 b. 5:2 c. 5:8 d. 4:7

Alex_Xolod [135]

The answer is A) 6:15, because if you divide both numbers by three, it'll give you 2:5

4 0

3 years ago

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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3 years ago

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devlian [24]

Answer:

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Step-by-step explanation:

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3 years ago

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Digiron [165]

Answer:

the answer is 2.9 x 10^0 and 3.2 x 10^-2

Step-by-step explanation:

10^0 is equal to 1 so 2.9 x 10^0 is really 2.9 times 1 which is less than 3

if i could get brainliest answer that would be great!

10^-2 is a exponential that makes a number go down so 3.2 going down is less than 3

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3 years ago

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