Answer:
Restrategize on means to be setting out a little proportion for clearing your bill and also making provision for handling the project ahead at same time.
4x5x2 ...small box 40cm3
10x4x8 ...big box 320cm3
Add both results
360cm3
Hope it works!
Good luck!
gen. Equation is...
x²/a² - y²/b² =1
a=length from origin to vertex
a=2
b=length from origin to focus
b=9
so equation is
x²/2² - y²/9² =1
-> (x²/4) - (y² /81) =1
Hope it helps...
Regards,
Leukonov/Olegion.
Answer:
x^2 +y^2
Step-by-step explanation:
Dude this is Pythragorean Theorem =)).
You may have missed this part of your lesson. It's ok and also not ok at the same time beacause pythagorean theorem is a meme.
Ok I'll explain once =)).
Thousands of years ago, a mathematician named Pythagore found out that if we have 3 squares, with one square's area equals to the sum of two other squares, then when we put the sides of the squares together to form a triangle , it'll always be a right triangle ( a triangle with one of its angle is 90°). A square area is measured by the length of its side multipled by itself. So he came up with the statement : In a right triangle, the length of the hypotenuse( the side that does not connect to the right angle) is equalled to the sum of squared other sides.
EG.
In this triangle, the right angle is ^ACB. If length of BC is <em>a</em><em>,</em><em> </em>length of AB is <em>c</em>, length of BC is <em>a</em>
A
I We have a formula :
I \ <em>a</em>^2 + <em>b</em>^2 = <em>c</em>^2
I \
I \
I <em>b</em> \ <em>c</em>
I \
I \
I______\
C <em> a </em> B
So now you can use pythagorean theorem to show off =)).
HOPE YOU LEARN WITH JOY AND HIGH GRADES !!!
Answer:
Step-by-step explanation:
Given is a function as 
Equating to 0 we have equation
If the function f(x) has x intercepts then the solutions are real
Let us use remainder theorem and change of signs rule
f(0) = 1>0
f(-1) = -1+3+1=3
f(-2) = -32+6+1<0
This implies there is a real root between -1 and -2.
f(1) = -1
Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.
f(2) = 32-6+1>0
Since f(1) and f(2) have different signs there exists a real root between 1 and 2.
Thus there are definitely three real solutions as
one between -1 and 0, one between 0 and 1, and third between 1 and 2.
