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Naily [24]
2 years ago
15

49-c< or equal to 45thanks

Mathematics
1 answer:
m_a_m_a [10]2 years ago
3 0

49-c\leq45\qquad\text{subtract 49 from both sides}\\\\-c\leq-4\qquad\text{change the signs}\\\\\boxed{c\geq4}

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Find (f-g)(x) if f(x)=3x-5 and g(x)=x2-5x+4
solmaris [256]

Answer:

Step-by-step explanation:

Hello,

(f-g)(x)=f(x)-g(x)=3x-5-(x^2-5x+4)\\\\=3x-5-x^2+5x-4\\\\=-x^2+8x-9

Thanks

4 0
3 years ago
If -48 equals 6(a) then a is what<br>​
Oksi-84 [34.3K]

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

<em>Expression:</em>

\frac{-48}{6}=\frac{6}{6}a

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3 years ago
(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}

1.

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

2.

q^0=1

3.

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

4.

(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}

5.

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

6.

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

7.

\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}

8.

(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}

8 0
3 years ago
I don't understand this question. Could someone help me? The graph is what is being looked at. I've already tested if they are o
Alex Ar [27]

The answer should be obtuse. I'm not sure why it is saying it's incorrect but none of the other answers make sense.

5 0
2 years ago
-7,5 + 4,2 =?<br><br><br> -0,9 * 2,7 =?
Gekata [30.6K]

Answer:

-7.5+4.2=-3.3

-0.9*2.7=-2.43

7 0
2 years ago
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