Answer:
12n + 60
General Formulas and Concepts:
<u>Pre-Algebra</u>
Distributive Property
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
(n + 12) × 5 + 7n
<u>Step 2: Simplify</u>
- Distribute 5: 5(n) + 5(12) + 7n
- Multiply: 5n + 60 + 7n
- Combine like terms: 12n + 60
9514 1404 393
Answer:
maximum difference is 38 at x = -3
Step-by-step explanation:
This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.
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Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.
h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2
Then the derivative is ...
h'(x) = 3x² +4x -15 = (x +3)(3x -5)
This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.
The maximum difference between f(x) and g(x) is 38 at x = -3.
Answer:
5/8
Step-by-step explanation:
12+30+20+10 +8 =80 total marbles
12+30+8=50 red blue and black
50/80
5/8
Answer:
For Lin's answer
Step-by-step explanation:
When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.
Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.
Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.
