All categories

3 years ago

What is 2+2 I have no idea please help

Mathematics

2 answers:

salantis [7]3 years ago

5 0

Answer:

2+2=4

Step-by-step explanation:

So you take the 2 numbers 2+2 and you use your finger counting to get to 4. It’s really hard mathematics but you’ll get there. Have a great day!

erastova [34]3 years ago

3 0

Answer:

2+2=4

Step-by-step explanation:

2+2=4

good question like if you have two chocolates and your mom gives you two more than there will be 4 and they will be too delicious

i hope this will help you :)

have a great day

You might be interested in

Fun question: Which Disney character is your favorite?

Vlada [557]

Answer:

Belle is my fav.

Step-by-step explanation:

She loves to read books like me and I hate people who always go out there and try to impress me with something.

5 0

3 years ago

How do I write 1,409/1,000 as a percentage?

Step2247 [10]

The answer should be 140.9%

5 0

3 years ago

2х + 6 = 7x - 14<br> What’s the answer?

Radda [10]

Answer:

x = 4

Step-by-step explanation:

2x + 6 = 7x - 14

2x - 7x = - 14 - 6

-5x = -20

x = 4

5 0

3 years ago

Read 2 more answers

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago