<span>A = 2b + ph
Where 'b' is the area of the base, 'p' is the perimeter of the base, and 'h' is the height.
Plug them inA = 2(24) + (22)(5)
SimplifyAnswer=158
</span>
Answer:
3 = 21 degrees
5 = 21 degrees
1 = 60 degrees
2 = 39 degrees
4 = 39 degrees
Step-by-step explanation:
This problem is really quite easy once you break it down. I assume you want an answer based on the attachment. If a triangle is Isosceles, then its bottom two angles are congruent, so angle 3 = angle 5.
Then, 138 degrees + x * 2 = 180 degrees.
x * 2 = 42 degrees
x = 21 degrees
So 3 and 5 are each 21 degrees
If a triangle is equilateral, then it is also equiangular, meaning each of its angles are equal to each other, and they are each 60 degrees. Then, we know angle 1 is 60 degrees. For 2 and 4, we want to take the measure of the combination of 2 and 3 (the full 60 degrees), and subtract it by the measure of angle 3 to get the measure of angle 2. 60 - 21 = 39 degrees. Angle 2 is 39 degrees. The same can be done with angle 5 and 4 because they are the same measurements.
<span><em>12 pennies, 3 nickles, and 2 dimes</em>
p = number of pennies
n = number of nickles
d = number of dimes
p(1) + n(5) + d(10) = 47
that is, the number of pennies x 1 cent + number nickles x 5 cents
+ number of dimes x ten cents equals 47 cents
p = 4n
p + n + d = 17
Substituting 4n for p in the above
4n + n + d = 17
5n + d = 17
Subtract 5n from each side
d = 17 - 5n
We will now substitute 4n for p and ( 17-5n ) for d in
the equation
p(1) + n(5) + d(10) = 47
4n(1) +n(5) + (17-5n)(10) = 47
9n + 170 - 50n = 47
-41n + 170 = 47
Subtract 170 from each side
-41n = 47 - 170
-41n = -123
Divide each side by -41
n = 3
Since p = 4n
p = 4(3)
p = 12
Since p + n + d = 17
12 + 3 + d = 17
15 + d = 17
d = 2
So we have 12 pennies, 3 nickles and 2 dimes
12 + 3(5) + 2(10) ?= 47
12 + 15 + 20 ?= 47</span>
Answer:
The root of the equation 
is x ≈ 0.162035
Step-by-step explanation:
To find the roots of the equation you can use the Newton-Raphson method.
It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess 
for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.
This is the expression that we need to use
For the information given:
For the initial value you can choose 
although you can choose any value that you want.
So for approximation 
Next, with 
you put it into the equation
, you can see that this value is close to 0 but we need to refine our solution.
For approximation 
Again we put 
into the equation
this value is close to 0 but again we need to refine our solution.
We can summarize this process in the following table.
The approximation 
gives you the root of the equation.
When you plot the equation you find that only have one real root and is approximate to the value found.
