All categories

3 years ago

Can someone please help on this problem and how it is done? Picture is attached. Thank you

Mathematics

1 answer:

Vikki [24]3 years ago

5 0

The angle passes 90 degrees plus an additional 55 degrees
m

You might be interested in

N= A-b <br> solve for A <br> hurry plzzzzz

Scrat [10]

Answer:

A=B+n

Step-by-step explanation:

plz mark me as brainliest

4 0

3 years ago

Read 2 more answers

Chelsea spent half of her weekly

Anvisha [2.4K]

Answer:

I believe it would be 16

Step-by-step explanation:

8 is half of 16 and if she had 8 dollars after spending half her allowance, she would've had 16 as her allowance

Final answer: Her allowance would be 16 dollars

3 0

3 years ago

Evaluate the expression shown below and write your answer as a fraction in simplest form.

Vesna [10]

Answer:

39/80

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Pls someone help I don’t get this

MakcuM [25]

Answer:

its 8

Step-by-step explanation:

first divide 16 by pi which is 3.14

then divide your answer by 2 after that.

5 0

3 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago

Can someone please help on this problem and how it is done? Picture is attached. Thank you​

Can someone please help on this problem and how it is done? Picture is attached. Thank you