Answer:
7(3^2) + 2(3).
That equals to 7(9) + 2(3).
That equals to 63 + 6.
That finally equals to 69.
Step-by-step explanation:
(C) 6 + 3√3
<u>Explanation:</u>
Area of the square = 3
a X a = 3
a² = 3
a = √3
Therefore, QR, RS, SP, PQ = √3
ΔBAC ≅ ΔBQR
Therefore,
In ΔBAC, BA = AC = BC because the triangle is equilateral
So,
BQ = √3
So, BQ, QR, BR = √3 (equilateral triangle)
Let AP and SC be a
So, AQ and RC will be 2a
In ΔAPQ,
(AP)² + (QP)² = (AQ)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
Similarly, in ΔRSC
(SC)² + (RS)² = (RC)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
So, AP and SC = 1
and AQ and RC = 2 X 1 = 2
Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR
Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3
Perimeter = 6 + 3√3
Therefore, the perimeter of the triangle is 6 + 3√3
Answer:
where a>0.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
Step-by-step explanation:
The real zeros are the result of setting each factor of the polynomial to zero. By reversing this process, we find:
- zero 1/2 is factor (2x-1)
We write them together with an unknown leading coefficient a which is negative so -a.
where a>0
The leading coefficient of a polynomial determines the direction of the graph's end behavior.
- A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.
- A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.
- This graph has all odd multiplicity. The graph will cross through the x-axis each time at its real zeros.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
Answer:
30 grams
Step-by-step explanation:
So, first I had to figure out what 1 tin weight. Which was 350 grams because I divide 5 into 1750. Then I did 350+4 which gave me 1,400. I knew that there was three packets so 30 grams per packet should be the answer because 30×3 is 90 grams.