Ask question

Ask question

All categories

3 years ago

11

Subtract 11 1/3 - 8 4/5

1 answer:

Hitman42 [59]3 years ago

4 0

$11 \frac{1}{3} - 8 \frac{4}{5} \\ \frac{34}{3} - \frac{44}{5} \\ \frac{38}{15}$

<h2>Solved ◇</h2>

You might be interested in

Help ?!-10+3x=8 <br> X=??

jekas [21]

Ok, so first off you need to remember that your goal is to get x alone.
How do you do this? Well, let's get rid of that pesky negative ten, shall we?
The opposite of a negative is positive, so you can move negative ten to the other side of the problem by adding ten. Make sure that you do operations on both sides. Once you add, you have 3x=18. When a number is right next door to an x, it means it is being multiplied. To get rid of the three, divide by three on both sides. Now your get x=6, which is your answer. Hopefully, I explained this well! :D

8 0

3 years ago

Read 2 more answers

Convert 35 pounds of candy to stones. (1 stone = 14 pounds)

bazaltina [42]

Answer:

2.4 stone

Step-by-step explanation: i think i did it in my head

7 0

3 years ago

Read 2 more answers

Is 2.99 irrational or rational?

ss7ja [257]

Answer:

rational

Step-by-step explanation:

its decimal expansion terminates after 9 and it can also be represented as fraction.

8 0

3 years ago

Read 2 more answers

PLEASE HELP Given: △KLM LM=12, m∠K=60°, m∠M=45° Find: Perimeter of △KLM.

worty [1.4K]

We have to find the perimeter of the triangle KLM.

We have been given that the length of the side LM=12, $m\angleK=60^\circ$ , and $m\angle M= 45^\circ$

Refer the attached image.

In a triangle sum of three angles should be $180^\circ$ .

So,

$m\angle K+m\angle L+m\angle M=180^\circ$

Plugging the values of angle K and angle M, we get:

$60^\circ+m\angle L+45^\circ=180^\circ$

So,

$m\angle L=180^\circ-105^\circ=75^\circ$

Now, that we have the measure of angle L, we will apply sine rule to find the length of the sides KL and KM.

Using the sine law for the triangle KLM, we get:

$\frac{sin K}{LM}=\frac{sin L}{KM}=\frac{sin M}{KL}$

Refer the image. Plugging the value of the sides of the triangle KLM and the angles of the triangle KLM, we get:

$\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}=\frac{sin 45^\circ}{x}$

Now using,

$\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}$

We get the value of 'y'

$y=\frac{sin 75^\circ}{sin 60^\circ} \times 12=\frac{0.9659}{0.866} \times 12=13.38$

So the length of the side KM is 13.38 units.

Now using,

$\frac {sin 60^\circ}{12}=\frac{sin 45^\circ}{x}$

We get the value of 'x'

$x=\frac{sin 45^\circ}{sin 60^\circ} \times 12=\frac{0.707}{0.866} \times 12=9.79$

So the length of the side KL is 9.79 units.

Now, to find the perimeter of the triangle KLM we need to sum up the length of the sides of the triangle KLM.

The perimeter of the triangle KLM $= KL+ LM + KM = 9.79 + 12 + 13.38 = 35.17$ units

6 0

3 years ago

Read 2 more answers

Does anyone get this problem on finding angles? (20 pts)

shusha [124]

A straight line is 180. To find the angle of b, you minus 39 from 180.

180-39=141

The answer is b=141

3 0

3 years ago

Read 2 more answers