Answer : 4 times
Here it's given that ,
- The height and base of the butterfly sitting on the stem (red butterfly) is two times greater than the height and base of the butterfly sitting on the flower .
And we need to find out how many times the area of red winged butterfly is greater than that of sitting on the flower (blue butterfly) .
Let us take ,
- base of blue butterfly be b
- height of blue butterfly be h
- Area be A .
Then ,
- base of red butterfly will be 2b .
- height of red butterfly will be 2h .
- Area be A' .
We know that ,
→ area of the triangle = 1/2 × base × height
So that ,
→ A/A' = (1/2 * b * h) ÷ (1/2 *2b *2h)
→ A/A' = bh/4bh
→ A/A' = 1/4
→ A' = 4A
<u>Henceforth</u><u> the</u><u> area</u><u> of</u><u> </u><u>blue</u><u> butterfly</u><u> is</u><u> </u><u>4</u><u> </u><u>times </u><u>greater</u><u> than</u><u> </u><u>that</u><u> of</u><u> </u><u>red </u><u>winged</u><u> butterfly</u><u> </u><u>.</u>
I hope this helps.
Answer:
p = -1 q = -4
Step-by-step explanation:
a system of eq and solve for p and q ??? can do :)
Eq. 1) 8p + 2q = - 16
Eq. 2) 2p - q = 2
use Eq .2 and solve for q
2p - 2 = q
plug into Eq.1 with q
8p +2(2p - 2) = - 16
8p +4p -4 = -16
12p = - 12
p = -1
plug -1 into Eq. 1 for p and solve for q
8(-1) + 2q = - 16
-8 + 2q = - 16
2q = -8
q = -4
Answer:
I don't know if you wanted it expanded using "Pascal's Triangle" or "Binomial Theorem" but either way is the same. Your expanded equation is:
a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5
I hope this is any help to you, have a nice day.
Step-by-step explanation:
We can start by getting rid of parenthesis:
4x-3x+6=21
Then we can combine like terms:
x=21-6
x=15
So our end product is x=15.
Hope I helped soz if I'm wrong ouo.
~Potato.
Copyright Potato 2019.
