6 pairs of shorts, for a
"s" dollars. A blazer costs 3 times as much, she bought one. The total was 139.50
6s + 3(6s) = 139.50
6s + 18s = 139.50
25s = 139.50 Divide everything by 25,
s = 5.58, which is how much the shorts were. Now for the blazers ;
It said they were three times the cost of shorts, if shorts are 5.58, it is 3*5.58.
3s
3(5.58)
16.74
The blazer cost 16.74
Equation: <span>3b+4b^2-a^3
Given: </span><span>a=3 & b=-5
Plug 3 in for (a) and -5 in for (b):
</span><span>3(-5)+4(-5)^2-(3)^3
</span>
Now follow PEMDAS:
<span>3(-5) + 4(-5)^2 - (3)^3
</span>-15 + 4(-5)^2 - (3)^3
-15 + 100 - (3)^3
-15 + 100 - 27
85 - 27
58
the asnwer is 58.
We have that
<span>question 1
Add or subtract.
4m2 − 10m3 − 3m2 + 20m3
=(4m2-3m2)+(20m3-10m3)
=m2+10m3
the answer is the option
</span><span>B: m2 + 10m3
</span><span>Question 2:
Subtract. (9a3 + 6a2 − a) − (a3 + 6a − 3)
=(9a3-a3)+(6a2)+(-a-6a)+(-3)
=8a3+6a2-7a-3
the answer is the option
</span><span>B: 8a3 + 6a2 − 7a + 3
</span><span>Question 3:
A company distributes its product by train and by truck. The cost of distributing by train can be modeled as −0.06x2 + 35x − 135, and the cost of distributing by truck can be modeled as −0.03x2 + 29x − 165, where x is the number of tons of product distributed. Write a polynomial that represents the difference between the cost of distributing by train and the cost of distributing by truck.
we have that
[</span>the cost of distributing by train]-[the cost of distributing by truck]
=[−0.06x2 + 35x − 135]-[−0.03x2 + 29x − 165]
<span>=(-0.06x2+0.03x2)+(35x-29x)+(-135+165)
=-0.03x2+6x+30
the answer is the option
</span><span>C: −0.03x2 + 6x + 30
</span><span>
</span>
Answer:
I think h is 20 I am not sure
