We want to find an integer x such that
2/3 < x/5 < 6/7
The steps and final answer to the problem are shown in the attached image below. The final answer is x = 4. Feel free to ask any questions you may have about any of the steps.
Answer: Ina fraction it could just be 2/9 in a decimal it is 0.22222222222
Step-by-step explanation: with the fraction 2 divided by nine is 2/9 simple as that. When you have a nine like that in the denominator take whatever number is the numerator, put 0 and a decimal point, and then just repeatedly press whatever the numerator is. Except for 9/9 then you just say 1. Hope this helped.
Answer:
See below.
Step-by-step explanation:
7.)
Supplementary angles add up to 180°. ∠B's supplement is 22°. This means ∠B is 158°. Since ∠A ≅ ∠B, the m∠A = 158°.
8.)
Again, supplementary angles add up to 180°. ∠P is a right angle meaning it is 90°. Therefore, the m∠Q is 90° as well.
9.)
Complementary angles add up to 90°. In order for ∠S and ∠T to be complementary, they both need to angle measurements less than 90° meaning they both need to be acute angles.
Now that we know that ∠T is acute, the only way ∠U can be its supplement is if it has an angle measure greater than 90°. Therefore, ∠U is an obtuse angle.
10.)
Similar to question 9, if an angle is obtuse, its supplement must be an acute angle.
Nadine does not use (variables).
The operations Nadine uses are the (same as) the operations Shay uses.
They use the operations in (same) order.
1.)
=(x-8i)(x+8i)
x^2+8ix-8ix-64i^2
x^2-64i^2
x^2-64(-1)
x^2+64
2.)
=(4x-7i)(4x+7i)
16x^2+28ix-28ix-49i^2
16x^2-49i^2
16x^2-49(-1)
16x^2+49
3.)
=(x+9i)(x+9i)
x^2+9ix+9ix+81i^2
x^2+18ix+81(-1)
x^2+18ix-81
4.)
=(x-2i)(x-2i)
x^2-2ix-2ix+4i^2
x^2-4ix+4(-1)
x^2-4ix-4
5.)
=[x+(3+5i)]^2
(x+5i+3)^2
(x+5i+3)(x+5i+3)
x^2+5ix+3x+5ix+25i^2+15i+3x+15i+9
x^2+6x+10ix+30i+25i^2+9
x^2+6x+10ix+30i+25(-1)+9
x^2+6x+10ix+30i-25+9
x^2+6x+10ix+30i-16
Hope this helps :)