The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11 
⇒x=11t
Area of square=
Area of square=
{as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
Learn more about The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11
⇒x=11t
Area of square=
Area of square={as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
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Answer:
butter brick and clay sink
oak and wheat float
Step-by-step explanation:
Y/x = 9/12 = 3/4
so
y = (3/4)x . . . . . . both x and y in dollars.
If the equation of the circle is x^2+ y^2 = 41, we must first understand the parts of the equation.
A general circle's equation is (x-h)^2+(y-k)^2= r^2
(h.k) is the radius of the circle
r is the radius of the circle
Another useful fact to know is that tangent lines touch the circle at one point (4,5)
Since in our original equation there are no h or k values, we can assume that the center of the circle is (0,0).
The formula for slope is <u>Y1-Y2</u>
X1-X2
We can break this down with our two points (center and tangent)
(0,0) and (-4,-5)
(X1,Y1) and (X2,Y2)
therefore, we will put the equation as such
<u>0-(-5)= 5</u> = <em> </em><u><em>5</em></u>
0-(-4)= 4 <em> 4</em>
<em>this is our slope from the center to the point of tangency.</em>
We know that tangent lines are perpendicular to the radius, which we've already found the slope of. Perpendicular lines are opposite reciprocals of the line they are perpendicular to.
Therefore, we take our slope from center to the tangent, and make it opposite and then take the reciprocal of that slope, which will give us the slope of the tangent line itself. (note: reciprocal means flip the numerator and denominator)
<u>5</u> = <u>-5</u> = <u>-4</u><u>
</u>4 4 5
Now, we have a point on the line, and the line's slope. We can use slope-intercept equation to find the equation of the line.
Slope-int y=mx+b
(x,y) is a point,
m is the slope
b is the y intercept ( the point where x=0, or where its on the y axis)
now we plug things in
(-4,-5) is our point,
<u>-4</u> is our slope
5
-5=<u>-4</u>(-4)+b After we plug things in, solve for b
5
-5= 3.2+b
-1.8= b or b= <u />1 <u>4</u>
5
Now we just need to rewrite our equation with all our components.
(-4.-5) = point
<u>-4</u> = slope<u>
</u>5
1 <u>4</u> = y-intercept<u>
</u> 5
<em>y=</em><u><em>-4</em></u><em> x+ 1 </em><u><em>4</em></u><em> This is the equation of the tangent line</em><u>
</u><em> 5 5</em>
Hope that helped
