Can someoene help with this please?

Mathematics

1 answer:

Natalija [7]3 years ago

6 0

Put them all in an equation where (m<1)+(m<2)+(m<3)=180 because all the angles of a triangle added together equals 180. From there, solve for x using algebra. Combine like terms and then isolate x. For the second part, use the value you found for x and find the measurement of angle 3. angle 4 is supplemental to angle 3 because angle 3 plus angle 4 is equal to 180 degrees. subtract the measurement of angle 3 from 180 to find angle 4. let me know if you have any trouble. i will find a computer to better explain if needed. im using a mobile phone, so communication is a bit limited

You might be interested in

A document that is 35 in. By 40 in. Is redrawn at 1 1/2 scale and then redrawn at 1/4 scale. What are the final dimensions of th

Andru [333]

The dimension of the document is 35 inches by 40 inches.
It is redrawn at a scale of 1 1/2 or 3/2 or 1.5
The dimension will be:
35 * 1.5 = 52.5 in
40 * 1.5 = 60 in

Then redrawn again at 1/4 or 0.25
52.5 * 0.25 = 13.125 in
60 * 0.25 = 15 in

So the final dimensions of the drawing is 13.125 in by 15 in

5 0

3 years ago

6+2[x+8]+3x+11+x help plzzzzz

Alex787 [66]

6+2(x+8)+3x+11+x
First you have to expand the brackets:
6+2x+16+3x+11+x
Now you group like terms (terms with the same or no variable):
2x+3x+x+11+16+6
And finally, combine like terms:
6x+33

Hope this helps :)

4 0

3 years ago

The perimeter of GHJ

Dafna1 [17]

They have a scale factor of 5:6 so first we need to find the perimeter of KLM:
$11 + 14 + 17 = 42$
Then apply the scale factor of 5:6:
$\frac{5}{6} = \frac{x}{42}$
Set the denominators equal:
$\frac{35}{42} = \frac{x}{42}$
So we see that the perimeter of GHJ is 35.

6 0

3 years ago

Subtract 1.05 from a certain number. Multiply the difference by 0.8, add 2.84 to the product then divide the sum by 0.01 and get

attashe74 [19]

Just do the whole process backwards
700*0.01=7
7-2.84=4.16
4.16/0.8=5.2
5.2+1.05=6.25

8 0

3 years ago

Simplify (8m2+m+7m2)-(3m-4m3+2m2)

kondaur [170]

$(8m^2+m+7m^2)-(3m-4m^3+2m^2)\\\\=8m^2+m+7m^2-3m+4m^3-2m^2\qquad|\text{combine like terms}\\\\=(4m^3)+(8m^2+7m^2-2m^2)+(m-3m)\\\\=4m^3+13m^2-2m$

4 0

3 years ago