Answer:
301.224 g
Step-by-step explanation:
Let's divide in the ratio 21:4 the volume of the medal (35cm^3)
copper = (21/25)(35cm^3) = 29.4cm^3
tin = (35 - 29.4)cm^3 = 5.6cm^3
<em>FOR THE COPPER</em>
8.84 : 1 = X : 29.4
X = 259.896(g)
<em>FOR THE TIN</em>
7.38 : 1 = X : 5.6
X = 41.328(g)
<em>(259.896 + 41.328)g </em><em>=</em><em> 301.224 g</em>
The first thing you must do is to calculate the weight of the gravel (w) that the truck holds (7.5 yards³). So, you have:
If the weight of 1 yard³ is 1.48 tons, then the weight of 7.5 yards³ is:
w=7.5x1.48 tons
w=11.1 tons
The problem says that the gravel will be placed in containers that each hold 3.7 tons of gravel,so, If they need 1 cotainer to place 3.7 tons, how many containers they need to place 11.1 tons?
1 container-------3.7 tons of gravel
x-------11.1 tons of gravel
x=(11.1x1)/3.7
x=11.1/3.7
x=3 containers
H<span>ow many containers of this size are needed to hold all the gravel from one truck?
</span>
The answer is: 3 containers.
Answer:
I assume you meant 4x^2
So the answer in simplified form is -4x^2-11x+3
Answer:
a) r(q) = 1000(q +1/q)
b) r'(q) = 1000(1 -1/q^2)
c) r'(10) = 990
Step-by-step explanation:
a) Revenue is the product of quantity and price:
r(q) = q·p(q) = q(1000(1 +1/q^2))
r(q) = 1000(q + 1/q)
__
b) The derivative is ...
r'(q) = 1000(1 -1/q^2)
__
c) The derivative evaluated for q=10 is ...
r'(10) = 1000(1 -1/10^2) = 1000(0.99)
r'(10) = 990
Cual numeros? No hay nada
