All categories

3 years ago

Number 8 please hurry

Mathematics

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

A rational number is normally number that can be expressed in the form of p/q where q ≠ 0. But, $\sqrt{42}$ cannot be expressed in the form of p/q. So. this means that

(Can I have Brainliest please? It really helps me help more people in Brainly!)

You might be interested in

NEED HELP ASAP

S_A_V [24]

Answer:

Step-by-step explanation:

8 0

2 years ago

Which rigid transformation(s) can map ABC onto FED

bogdanovich [222]

Answer:

I believe it would be B) Reflection, then translation

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the sum of all integers between 19 and 77 squared rooted?

Alex Ar [27]

4<√19<5
8<√77<9
so the quality numbers are 5,6,7,and 8
5+6+7+8=26
so the answer is 26

4 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

ddd [48]

Answer:

$f'(x) = 4x^2 - 3$ for $x \le -3$

Step-by-step explanation:

See attachment for proper question

Given

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

For

$x \ge -3$

Required

Determine the inverse function

$f(x) = -\frac{1}{2}\sqrt{x + 3}$

Replace f(x) with y

$y = -\frac{1}{2}\sqrt{x + 3}$

Swap the positions of x and y

$x = -\frac{1}{2}\sqrt{y + 3}$

Multiply both sides by -2

$-2 * x =-2 * -\frac{1}{2}\sqrt{y + 3}$

$-2x =\sqrt{y + 3}$

Square both sides

$(-2x)^2 =(\sqrt{y + 3})^2$

$4x^2 =y + 3$

Make y the subject

$y = 4x^2 - 3$

The inverse has been solved. So, we need to replace y with f'(x)

$f'(x) = 4x^2 - 3$

Next, is to determine the interval

$x \ge -3$

Change inequality to $\le$

$x \le -3$

Hence, the inverse function is:

$f'(x) = 4x^2 - 3$ for $x \le -3$

7 0

3 years ago

Samira and tineka wrote a 4-digit numbers

Vika [28.1K]

The value of the 8 in Samira's number is 8,000. The value of the 8 in Tineka's number is 800.

I believe your answer would be Option D) 1000.

Hope this helps! :)

4 0

3 years ago

Read 2 more answers