Someone pls help me with this, i would appreciate it. last math question i need help with

Kelvin finished $\frac{2}{3}$ 2 3 of his test in a $\frac{1}{2}$ 1 2 hour. Express this rate in hours per test.

svet-max [94.6K]

Complete Question

Kelvin finished 2/3 of his test in a 1/2 hour. Express this rate in hours per test.

Answer:

3/4 hours per test

Step-by-step explanation:

2/3 of his test = 1/2 hour

This rate in hours per test is calculated as:

x = 1/2 /2/3

x = 1/2 hour ÷ 2/3 test

x = 1/2 × 3/2

x = 3/4 hours per test

This rate in hours per test is given as 3/4 hours per test

4 0

3 years ago

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$

The length cannot be negative, so CD=8 cm and

$CA=\sqrt{3}CD=8\sqrt{3}\ cm$

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

$AB=2CA=16\sqrt{3}\ cm$

By the Pythagorean theorem,

$BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm$

3 0

3 years ago

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

Olegator [25]

Answer:

The length of shadow decrease at 0.48 m/s.

Step-by-step explanation:

The distance of man from building = x

Thus initially, x = 12

It is given that dx/dt= -1.3 m/s

Let the shadow height = Y

Now solve for the $dy/dt$ when x=4

from the diagram (triangle) it can be seen similar triangles with similar base/height ratios.

$\frac{2}{12-x} = \frac{y}{12} \\Y = \frac{24}{12 - x} \\ \frac{dy}{dt} = \frac{24}{(12-x)^{2}} \times \frac{dx}{dt} \\at x=4, dy/dt \ becomes: \\\frac{24}{(12-4)^2} \times -1.3 \\= -0.48 m/s$

8 0

3 years ago

The circumference (C) of a swimming pool is 47 feet. Which formula can you use to find the radius (r) if you know that C = 2πr ?

vodka [1.7K]

To find the formula to find radius, solve for r using the formula for circumference.

to solve:
1. divide each side by 2 $\pi$
2. you get r = c/2 $\pi$

5 0

3 years ago

Read 2 more answers

(07.07)

dangina [55]

Answer:

c

Step-by-step explanation:

1.58 times the number of n bags equal to z money nedded to buy n bags of popcorn

7 0

4 years ago