All categories

2 years ago

Is the square root of 4 plus the square root of 16 rational?

Mathematics

2 answers:

evablogger [386]2 years ago

8 0

Answer: Yes

Step-by-step explanation:

The square root of 4 + the square root of 16 = 6 which is a rational number.

andre [41]2 years ago

8 0

Answer:

Yes, it is rational

Step-by-step explanation:

sqrt(4) =±2

sqrt(16) =±4

Both of these numbers are rational, so the sum of rational numbers is rational

You might be interested in

Mathematical induction

jenyasd209 [6]

Answer:

Process to proof Theorem and Formula nrealted to natural number.......

7 0

2 years ago

You invest 1000 in an account at 2.5% per year simple interest how much will you hav in your account at the beginning of the 4th

Lina20 [59]

Each year + $25. 4 years = $100. 100+1000= $1100

4 0

3 years ago

HELP ASAPPPPPPPPP ILL MARK BRAINLIEST

pogonyaev

Answer:

the answer would be C

Step-by-step explanation:

the opposite of exponents is square root it

4 0

3 years ago

Read 2 more answers

Solve the inequality: a - 4 <10

harkovskaia [24]

<h3>Answer: a < 14 (Choice B)</h3>

==================================================

Explanation:

Add 4 to both sides

a-4 < 10

a-4+4 < 10+4

a < 14

The reason why we add 4 to both sides is to undo the "minus 4" that's happening to the variable.

5 0

2 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

2 years ago

Read 2 more answers