Answer:
16 ways
Step-by-step explanation:
Number of available options other than the base model = 4
Hence ;
Jacobsen can choose :
Base model OR (base model + one option) OR (base model + 2 options) OR (Base model + 3 options) OR (Base model + 4 options)
4C0 + 4C1 + 4C2 + 4C3 + 4C4
Recall:
nCr = n! / (n-r)!r!
Using calculator :
4C0 + 4C1 + 4C2 + 4C3 + 4C4
1 + 4 + 6 + 4 + 1
= 16 ways
Answer:
∠P = 39º
∠Q=120º
∠R = 21º
Step-by-step explanation:
m∠P+m∠Q+m∠R=180º (sum of ∠s in a triangle)
2x-3+6x-6+x=180º (substitution)
9x-9=180º (algebra)
9x=189º (algebra)
x=21º (algebra) (this is ∠R)
(2x-3)=39º (algebra) (this is ∠P)
6x-6=120º (algebra) (this is ∠Q)
There you go buddy, hope this helps
Thank you.
7^5 times 4^5 is equivalent to (7 times 4)^5, which is 28^5. You could either leave your answer in this format or evaluate 28^5: 17210368.
(153^2)^7 is best done using the rule of exponents (x^a)^b = x^(ab). Multiply the exponents 2 and 7 together: (153^2)^7 = 153^(2*7) = 153^14.
(10^5)(4^5) has two factors, both of which have the exponent 5. We can rewrite this expression as (10*4)^5, which equals 40^5. No, this is not equal to 14^5!