Answer:
This is possible.
Step-by-step explanation:
We can say that m<E=m<E, because of the Reflexive Property
Then, we have angles JKL and ELJ, which are equal through the peripheral angle theorem.
With these two angles, we can say that triangles ELK and EJL are similar, by the Angle-Angle Postulate (AA).
Then we can create this ratio through the Corresponding Parts of Similar Triangles Theorem, (CPST), 
.
With this ratio, we can cross multiply to get the desired result
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Hope this helps with your RSM problem
Yup, i caught ya.
The smallest value is $0.25 (a quarter) because you can use two quarters and a half dollar. Only one of them can't be a half dollar, but the other two can be.
The radii of the frustrum bases is 12
Step-by-step explanation:
In the figure attached below, ABC represents the cone cross-section while the BCDE represents frustum cross-section
As given in the figure radius and height of the cone are 9 and 12 respectively
Similarly, the height of the frustum is 4
Hence the height of the complete cone= 4+12= 16 (height of frustum+ height of cone)
We can see that ΔABC is similar to ΔADE
Using the similarity theorem
AC/AE=BC/DE
Substituting the values
12/16=9/DE
∴ DE= 16*9/12= 12
Hence the radii of the frustum is 12
Answer:
Step-by-step explanation:
Use synthetic division to answer this. If the remainder is zero, then we can safely assume the divisor (x + 7) is a factor of the polynomial f(x)= x^3-3x^2+2x-8.
We use -7 as the divisor in synth. div. This comes from the factor (x + 7):
-7 / 1 3 2 -8
-7 28 -210
-------------------------
1 -4 30 -218
Here, the remainder is -218, not zero, so no, (x+7) is not a factor of f(x)= x^3-3x^2+2x-8.
Step-by-step explanation:
step 1. a circle with r = 2 has area of (pi)r^2 = (pi)(4) = 4pi
step 2. a circle with r = 3 has area of (pi)r^2 = (pi)(9) = 9pi
step 3. the area, A, of a circle is half the sum of the area of the circles with r = 2 and r = 3
step 4. A = (4pi + 9pi)/2 = 13pi/2 = 6.5(pi).
step 5. the answer is not 6.5 but is 6.5(pi) or 6.5 times pi.
