What is the solution for 16=2/3f

Kristy makes 5$ per hour babysitting and 12$ per hour giving music lessons. One weekend, she worked a total of 18 hours and made

MariettaO [177]

Let Karen works x hours baby sitting
So, she worked (18-x) hours giving music lessons
So,
According to question
5x + 12(18-x)=139
This gives
7*x=77
x=11
So,
Karen worked 11 hours baby sitting and (18-11)=7 hours giving music lessons

3 0

3 years ago

Answer pls thanks kfdsdatttdfyui8u90g

diamong [38]

Answer:

Step-by-step explanation:

960 75%

6 0

3 years ago

PLEASE HELP! i have by the end of the day to turn this in. PLEASE DONT GUESS

ivann1987 [24]

Answer: See step by step

Step-by-step explanation: Angle 2 and 3 are vertical angles since they both share a vertex and has cross intersecting lines. Angle 6 and 7 are supplementary angles because they forma linear pair.

7 0

3 years ago

Read 2 more answers

The length of a rectangle is "a" inches and the width is 3 inches more than its length. find the perimeter and the area

Alex73 [517]

Answer:

Step-by-step explanation:

Givens

Length = a

Width = a + 3

Formulas

P = 2L + 2w

Area = L * w

Solution

Perimeter

P = 2L + 2w

P = 2*a + 2(a + 3) Remove the Brackets

p = 2a + 2a + 6 Combine

P = 4a + 6

Area

Area = L * w

Area = a (a + 3)

Area = a^2 + 3a

7 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago