Answer:
25 necklaces
Step-by-step explanation:
each necklace brings $16 profit (20-4=16)
400÷ 16= 25
so, at 25 necklaces, there will be enough profit to break even after paying for the advertising.
Answer: x = {-1, -3, 2}
<u>Step-by-step explanation:</u>
x³ + 2x² - 5x - 6 = 0
Use the rational root theorem to find the possible roots: ±1, ±2, ±3, ±6
Use Long division, Synthetic division, or plug them into the equation to see which root(s) work <em>(result in a remainder of zero)</em>.
I will use Synthetic division. Let's try x = 1
1 | 1 2 -5 -6
|<u> ↓ 1 3 -2 </u>
1 3 -2 -8 ← remainder ≠ 0 so x = 1 is NOT a root
Let's try x = -1
- 1 | 1 2 -5 -6
|<u> ↓ -1 -1 6 </u>
1 1 -6 0 ← remainder = 0 so x = -1 is a root!
The coefficients of the reduced polynomial are: 1, 1, -6 --> x² + x - 6
Factor: x² + x - 6
(x + 3)(x - 2)
Set those factors equal to zero to solve for x:
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
Using Synthetic Division and Factoring the reduced polynomial, we found
x = -1, -3, and 2
- Length (l) = 45 m
- Breadth (b) = 30 m
- We know, perimeter of a rectangle = 2(length + breadth)
- Therefore, perimeter of the rectangle
- = 2(I + b)
- = 2(45 + 30) m
- = 2 × 75 m
- = 150 m
- So, the distance travelled by Rubina
- = 4 × 150 m
- = 600 m
<u>Answer:</u>
<em><u>The </u></em><em><u>distance </u></em><em><u>travelled</u></em><em><u> </u></em><em><u>by </u></em><em><u>Rubina </u></em><em><u>is </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>m.</u></em>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
Ok thats really cool thanks :)
Answer:
see below
Step-by-step explanation:
(ab)^n=a^n * b^n
We need to show that it is true for n=1
assuming that it is true for n = k;
(ab)^n=a^n * b^n
( ab) ^1 = a^1 * b^1
ab = a * b
ab = ab
Then we need to show that it is true for n = ( k+1)
or (ab)^(k+1)=a^( k+1) * b^( k+1)
Starting with
(ab)^k=a^k * b^k given
Multiply each side by ab
ab * (ab)^k= ab *a^k * b^k
( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)
Therefore, the rule is true for every natural number n