What is the fifth term of an arithmetic sequence whose first term is 40 and whose seventh term is 121?

94

Step-by-step explanation:

Fifth term= a+4d

a=40

a+6d=121

using elimination method

-6d = -81

so, d= 13.5

Therefore; a+4d

= 40+4(13.5)

= 94.

Therefore, fifth term is 94...

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°

if you are looking for the unit rate then the answer would be 9 because unit rate is distance divided by time.

step-by-step explanation: