$\tan (m\angle H) = \frac{IJ}{JH} \\ \\ \implies \: \tan (m\angle H) = \frac{ \sqrt{10} }{ \sqrt{30} } \\ \\\implies \: \tan (m\angle H) = \frac{{1} }{ \sqrt{3} } \\ \\ \implies \: \tan (m\angle H) = \tan 30 \degree \\ \\ \implies \: m\angle H = \boxed{30} \degree$

6 0

2 years ago

PLEASEE HELPP MEEEEE !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ksju [112]

G(x)= $x^{2}$ +2

7 0

2 years ago

Read 2 more answers

What are the solutions to the quadratic equation -2x^2 + 6x + 3 = 0?

mario62 [17]

For this, we will be using the quadratic formula, which is $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$ , with a=x^2 coefficient, b=x coefficient, and c = constant. Our equation will look like this: $x=\frac{-6+/-\sqrt{6^2-4*(-2)*3}}{2*(-2)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-6+/-\sqrt{36+24}}{-4}$

Next, do the addition: $x=\frac{-6+/-\sqrt{60}}{-4}$

Next, your equation will be split into two: $x=\frac{-6+\sqrt{60}}{-4},\frac{-6-\sqrt{60}}{-4}$ . Solve them separately, and your answer will be $x=-0.436,3.436$

5 0

3 years ago