Helpfind the equation of the linear function q(16)=1 that is parralel to f(x)=1/3x+39 It wants it in slope intercept form

Step-by-step explanation:

So you have (16, 1)

When two linear functions are parallel, they have the same slope.

So function q also has a slope of 1/3x. You have a set of points so you can plug them in a slope intercept equation to find b.

y= mx+b

m= slope

y= y-intercept

1= 1/3(16) +b

1 - (16/3) = b

b= -13/3

so then,

the first one, all those points on the line just make a strait line.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°