Does anyone know how to do equations in 6th-grade math? i will be rewarding 15 points to whoever helps me

Yea I can help you with these problems I’ve done this a long time ago and so far I remembered how to do this. So for number 1. X=4 in this problem so we need to substitute x into 4 and recreate this expression: 4+1=5, and yes it is a solution. As for number 2, w would equal to 2 so 13-2=10 isn’t a solution because 13-12 is actually 11, so this is a false. Now number 3, v would equal to 10 so you’d have to replace 10 into this expression; so it’s going to be 2(10)=12 which isn’t a solution because 2 times 10 equals to 20. Now for number 4, p would have to equal to 7, here is the expression; 14/7=2, is a solution because if you’re going to check it do 7(2) and it would actually equal to 14 so yes it is a solution. Now for number 5 you would have to equal w to 3 so do this in an expression; 8+3=11, and yes that is a solution because you can check it as 11-3 which would equal to 8. Now for number 6 t would have to equal to 5 so replace t into 5 which the expression is going to be; 4(5)=20 so yes it is a solution because you can check it by doing 20/5=4. Now for number 7 s would have to equal to 3 so in this expression would have to be; 12/3=4 and yes that is a solution because you can check it by doing 3 times 4 which would equal to 12. Now for the last problem 8, d would have to equal to 8 so you replace d into 8 which the expression is going to be; 6+8=14, it’s not going to be 6+8=15 because it doesn’t equal to that since if you want to check the right answer just subtract 14 by 8 and you would get 6. So the last problem isn’t going to be a solution.

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )