Answer:
x = 112
Step-by-step explanation:
We can create two equations 1. one that calculates calories burned and 2.another that calculates minutes:
20*x + 5*y = 1200
x + y = 120
Then we can solve for x
x = 120 - y
then we can plug that into the top equation:
20(120 - y) + 5y = 1200
divide by 5:
4(120 - y) + y = 240
distribute:
480 - 4y + y = 240
and then do all that combining like factors what not and simplify and solve:
480 - 3y = 240
-3y = -240
y = 8
NOW WE CAN JUST PLUG IN THE ANSWER FOR Y TO FIND X
8+x =120
x = 112
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
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<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
Answer:
$3.94
Step-by-step explanation:
You will need to use the compound interest formula for this.

P = initial balance
r = interest rate
n = number of times compounded annually
t = time
Your equation will look like this:
= 3.94
Step-by-step explanation:
Hi there!
Given;
= 5(π\6)
We have;
π = 180°
Keeping value of π in the question;
= 5(180°/6)
= 5*30°
= 150°
Therefore, answer is option B.
<u>Hop</u><u>e</u><u> </u><u>it </u><u>helps</u><u>!</u>