Simplify the following expression, if you can see the picture the expression is as follows:

Liam is constructing a square in which two of its vertices are points A and B. He has already used his straightedge and compass

AfilCa [17]

Answer:

Step-by-step explanation:

B

3 0

2 years ago

Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of

Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2 are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

$v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j$

The unit vector in the direction of force will be

$\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}$

The magnitude of the force is 40lb, so the force will be

$F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)$

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

$v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k$

$\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}$

The magnitude of the force is 60lb, so the force will be

$F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)$

The resultant of the two forces is

$F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k$

The magnitude force will be

$|F|=\sqrt{(20\sqrt{2} )^2+(50\sqrt{2} )^2+(30\sqrt{2} )^2} \\\\=\sqrt{800+5000+1800} \\\\=\sqrt{3100} \\\\=55.68$

to (1 decimal place)=55.7lb

b) The direction angle of force F

The angle formed by F and x axis

$\alpha=\cos^{-1}(\frac{20\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.5080)\\\\=59.469$

The angle formed by F and y axis

$\alpha=\cos^{-1}(\frac{50\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(1.270)\\\\=$

The angle formed by F and z axis

$\alpha=\cos^{-1}(\frac{30\sqrt{2} }{\sqrt{3100} } )\\\\=\cos^{-1}(0.7620)\\\\=40.359$

8 0

2 years ago

Kali left school and traveled toward her friend’s house at an average speed of 40

ioda

Matt needs to travel 3 hours.
If 40km=1 hour
400km=? But there are two distances
200/40*1=5 hours
Therefore Kali needs to travel 5 hours to be 200 km away from the house.
If 50km=1 hour
200km=?
200/50*1hr=4hrs Therefore Matt needs to travel 4hrs to be 200 km away from the house. But Kali traveled 1hr earlier than Matt. 4hrs-1hr=3hrs
Therefore Matt has to travel for 3 hours to be 400km away from Kali.

3 0

2 years ago

A pizzeria charges $8 for a large cheese pizza, plus $2 for each topping. The total cost for a large pizza is given by the equat

Leni [432]

Slope is the rise and run of a point on the graph to the other
the y intercept is where a line crosses the y axis for example, (0,5) so 5 is the y intercept

(0,8), (1,10), (2, 12), (3, 14), (4, 16), (5, 18)

7 0

2 years ago

Read 2 more answers

A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min na

Oxana [17]

Answer:

(a) 2.29 km/h

(b) 9 km/h

Step-by-step explanation:

For part (a) you have to apply<em> the average speed formula</em>, which is defined by:

$v=\frac{d}{t}$

where d is the total distance traveled and t is the total time needed.

$d=\frac{4.00}{1.75}=2.29$ km/h

For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:

The nap time in hours is:

90/60 = 1.5 h (because there are 60 minutes in one hour)

The running time is:

T= 1.75 - 1.5 = 0.25 h

Let t1 represent the time before the nap and t2 the time after the nap:

t1+t2 = T

t1+t2 = 0.25

You have to apply the formula d=vt before and after the nap:

-Before the nap, the distance traveled was 0.50 km

0.50 = v1t1

-Afer the nap, the distance traveled was 3.50 km

3.50=v2t2

But v2=2v1 (because after the nap the rabbit runs twice as fast)

You have to solve the system of equations:

t1=0.25-t2 (I)

v1t1=0.50 (II)

2v1t2=3.50 (III)

Replacing (I) in (II)

v1(0.25-t2)=0.50

Applying distributive property and solving:

0.25v1-v1t2=.050

For (III) you have that v1t2=3.50/2=1.75. Hence:

0.25v1-1.75=0.50

Solving for v1:

v1 = 9 km/h

7 0

1 year ago