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What's up, my mammals? This is Sid the sloth from the movie "Ice Age". And we're about to do this new dance, the "Continental Dr

Bogdan [553]

Answer:

this question > any other question

Step-by-step explanation:

4 0

4 years ago

Hi, can someone help on this?

marissa [1.9K]

Answer:

i beleive its 9 and 1 if im reading that right

Step-by-step explanation:

8 0

3 years ago

Sales of $10,000 increase by 65% each year.

Sati [7]

Answer:

hope this helps

Step-by-step explanation:

65% of 10,000

.65 * 10,000

6500

After one year-

10,000 +6500

16500

7 0

4 years ago

How do you find the answer?5y+8=-6

sammy [17]

Answer:

Step-by-step explanation:

5y+8=-6

-8

5y=-14/5

y=-2.8

7 0

3 years ago

Read 2 more answers

Make 'h' as the subject. Remember to rationalize the denominator while solving.

Sliva [168]

Hello there BeGuiLE !

To solve your question, we must first bring 'h' to one side of the equation. Let's bring all the terms with the variable 'h' to the left side of the equation. So,
$\large{\sf\sqrt{ 3 } h = 1.6+h}$
→ $\large{\sf\sqrt{3}h-h=1.6 }$

Now, let's combine the 2 terms containing the variable 'h'.
$\large{\sf\sqrt{3}h-h=1.6 }$
→ $\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
$\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$
→ $\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
$\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }$
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }$
→ $\boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}$

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
$\large{\sf\:h=0.8(\sqrt{3}+1)}$
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ $\large{\sf\:h=0.8(1.73+1)}$
→ $\large{\sf\:h=0.8(2.73)}$
→ $\boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}$

I hope this will help you.