Number 11; What do you have to do to solve it?

Mathematics

1 answer:

taurus [48]3 years ago

3 0

very positive about these results

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Solve this for most Brainly

Ede4ka [16]

Answer:

1 - 22

2 -

$- 4x {y}^{3} - 6 {x}^{4} {y}^{3} + 7 {y}^{4} - 8 {x}^{4} {y}^{4}$

3- ( x = -4)

4 - Irrational numbers

$\sqrt{53} \: \: \: \: \: \sqrt{19} \: \: \: \: \pi$

integers

$0 \: \: \: \: \: \frac{ - 12}{6}$

rational

$\frac{2}{3} \: \: \: \: \frac{6}{45}$

natural numbers 8

5 0

3 years ago

Please help me if you can. Please also write how you got the answer thank you.

babunello [35]

Answer:

Question 9: Variables: (smallest) s, q, r (largest)

Question 10: 5 whole numbers (7, 8, 9, 10, and 11)

Step-by-step explanation:

For question nine, there are two given statements... s=q-2 and q<r. Say we plug in 10000 (a really big #) in for q, then we would get s=9998 and r>10000. This way, we can see that s would be the smallest, then q, and r is the largest. <em>(q<r can be written as r>q)</em>

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For question 10, it states $\frac{1}{4}$ . This can be split into $\frac{3}{x}$ and $\frac{3}{x}$ . When x is 12 in the first equation then $\frac{3}{12} = \frac{1}{4}$ and when x is 6 in the second equation $\frac{3}{6} =0.5$ (0.5 is also $\frac{1}{2}$ ). Therefore, x must be a whole number less than 12 and greater than 6, and it cannot be either 12 or 6. Whole numbers between 6 and 12 are 7, 8, 9, 10, and 11 or 5 whole numbers.