Questions in Grade High-school

Questions in Grade High-school

[Done] Man created tools and the first art close to the same time. A. True B. False

[Done] A person is standing exactly 36 ft from a telephone pole. There is a [tex]$30^{\circ}$[/tex] angle of elevation from the ground to the top of the pole. What is the height of the pole? A. 12 ft B. [tex]$12 \sqrt{3} ft$[/tex] C. 18 ft D. [tex]$18 \sqrt{2} ft$[/tex]

[Done] Complete the steps to simplify $\sqrt[5]{4} \cdot \sqrt{2}$. Rewrite using rational exponents. $\checkmark \quad 2^{\frac{2}{6}} \cdot 2^{\frac{1}{2}}$ $4^5 \cdot 2^2$ $4^{\frac{1}{6}} \cdot 2^{\frac{1}{4}}$ COMPLETE What is the least common denominator of the exponents? $\square$

[Done] POLYNOMIAL (P) OR NON-POLYNOMIAL (NP)? 1. 5 2. -2x^2 3. (3/4)x + 21 4. 5x^2 - 4xy + 2y^2 5. a^(-2) + 4 6. sqrt(9h) 7. sqrt(25) 8. (2/k) + 14 9. 1/4 10. 10 - b^(1/2)

[Done] Calculate the following and state your answer in scientific notation: \[\frac{(3 \times 10^2 \text{ kg})(4.0 \text{ km})}{5 \times 10^2 \text{ s}^2}\]

[Done] A shopkeeper buys 6 balloons for one rupee. How many balloons should he sell for one rupee so that he makes a profit of 20%? 1. 3 2. 4 3. 5 4. 6

[Done] Supply the correct modal to complete the sentence. Write your answer on your answer sheet. 1. You be exhausted after your business trip. Come in and have a drink. 2. Baggage be left unattended. 3. You see the Intramuros and Luneta Park when you go to the Philippines. 4. My doctor said that I stop smoking or I'll risk serious problems. 5. Tell your brother he __ use my car for the delivery of parcels.

[Done] In \(\triangle PQR\), \(|PO| = 15\) cm, \(|OR| = 16\) cm, \(|RP| = 9\) cm. The bisector of \(\angle P\) meets \(OR\) at \(A\) and \(M\) is the midpoint of \(OR\). Calculate \(|AM|\).

[Done] -x³ + 4x = mx Rearranged as: x³ - 4x + mx = 0 Factored: x(x² - 4 + m) = 0 Using the quadratic formula for x² - 4 + m = 0: x = \frac{0 \pm \sqrt{0 - 4(1)(m - 4)}}{2(1)} = \frac{\pm \sqrt{-4m + 16}}{2} = \pm \frac{\sqrt{4 - m}}{1} = \pm \sqrt{4 - m} Therefore, the solutions are: x = 0, \pm \sqrt{4 - m}

[Done] What is the general equation of a cosine function with an amplitude of 3, a period of [tex]$4 \pi$[/tex], and a horizontal shift of [tex]$-\pi$[/tex]? A. [tex]$y=4 \pi \cos (3(x-\pi))$[/tex] B. [tex]$y=3 \cos (4 \pi(x+\pi))$[/tex] C. [tex]$y=3 \cos (0.5(x+\pi))$[/tex] D. [tex]$y=4 \pi \cos (0.2(x+\pi))$[/tex]