How do you graph y=7

Mathematics

2 answers:

Damm [24]3 years ago

6 0

You go to x instead of y on a graph , find 7 and put a dot on it

zmey [24]3 years ago

4 0

I think the line y=7 would be the value 7 drawn across they-axis going horizontally

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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.