All categories

3 years ago

3x – 4y = -5 y = 4.3 - 2 Is (1, 2) a solution of the system? Choose 1 answer:

Mathematics

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

No. The correct answer is (1.4, 2.3)

Step-by-step explanation:

y= 4.3 - 2

y= 2.3

3x - 4y = -5

3x - 4(2.3) = -5

3x - 9.2 = -5

3x = -5 + 9.2

3x = 4.2

x = 4.2 ÷ 3

x = 1.4

(1.4 , 2.3)

You might be interested in

Evaluate 4log4(3,100)

alexdok [17]

Answer:

${ \tt{4 log_{4}(3100) }} \\ = { \tt{4 log_{4}(775 \times 4) }} \\ = { \tt{4 log_{4}(775) + 4 log_{4}(4) }} \\ = { \tt{4( (\frac{ log_{10}775 }{ log_{10}4 } ) +1)}} \\ = { \tt{4(4.8 + 1)}} \\ = { \tt{23.2}}$

8 0

3 years ago

What is x + 3 +2x =x + 3

algol13

X=0 bexause the 3s cancel out and your left with 2x=0 so x=0

8 0

3 years ago

Read 2 more answers

Can someone help me?

Vikki [24]

Answer:

x=7

Step-by-step explanation:

35÷5=7 :)))

8 0

3 years ago

Read 2 more answers

Can some help me with number 2?

andrezito [222]

Answer:

$54.85

Step-by-step explanation:

Total=$42.85

Total + Tax(8%)=$46.28

Total + Tax(8%) + Tip(20%)=$54.85

4 0

3 years ago

How do you use the limit comparison test on this particular series?<br>Calculus series tests

barxatty [35]

Compare $\dfrac1{\sqrt{n^2+1}}$ to $\dfrac1{\sqrt{n^2}}=\dfrac1n$ . Then in applying the LCT, we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1$

Because this limit is finite, both

$\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}$

and

$\displaystyle\sum_{n=1}^\infty\frac1n$

behave the same way. The second series diverges, so

$\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n$

is divergent.

4 0

4 years ago