Answer:
The angle Rick must kick the ball to score is an angle between the lines BX and BZY which is less than or equal to 32°
Step-by-step explanation:
The given measures of the of the angle formed by the tangent to the given circle at X and the secant passing through the circle at Z and Y are;
The direction Rick must kick the ball to score is therefore, between the lines BX and BXY
The angle between the lines BX and BXY = ∠XBZ = ∠XBY
The goal is an angle between 
Let 'θ' represent the angle Rick must kick the ball to score
Therefore the angle Rick must kick the ball to score is an angle less than or equal to ∠XBZ = ∠XBY
By the Angle Outside the Circle Theorem, we have;
The angle formed outside the circle = (1/2) × The difference of the arcs intercepted by the tangent and the secant
We get;
∠XBZ = (1/2) × (122° - 58°) = 32°
The angle Rick must kick the ball to score, θ = ∠XBZ ≤ 32°
To find the maximum height you need to find the vertex:(h,k)
Your equation is in vertex form a(x-h)+k and the vertex is (h,k) where k is the maximum height and the h is the distance it went to reach the maximum height.
k=6 so the kangaroo's maximum height is 6 feet.
To find how long is the kangaroo's jump, take a look at the graph. You will notice that the parabola ends at the distance the kangaroo jumped. You will also see that it is the one of the x-intercepts.
-.03(x-14)^2+6=0
-.03(x-14)^2+6-6=0-6
-.03(x-14)^2=-6
-.03/-.03(x-14)=-6/-.03
(x-14)^2=200
[(x-14)^2]^.5=200^.5
x-14=(200)^.5
x-14+14=(200)^.5+14
x≈28.14 feet
The kangaroo jumped a distance of 28.14 feet.
You will notice that the square root of a number gives you two solutions a positive and a negative one. The other solution is -.14, which we know distance is not negative so we do not use that solution. Also, I used the ^.5 instead of using the square root. It is the same.
Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
.
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
.
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
Answer:
There are many ways to show it like-
Step-by-step explanation:
2/5 as it’s already shown, or you can use 2 divide by 5, many people use 2[5] that’s the ways they showed my in my school idk if its different for you
I gotchu.
<span>3x-9x+6=20-3x+4x
Combine like terms.
-6x+6=20+x
Bring x over to the left side.
-x -x
-7x+6=20
bring 6 to the right side.
-6 -6
-7x=14
Divide -7 on both sides to isolate x.
x= -2
CHECK
</span><span>3x-9x+6=20-3x+4x
-6(-2)+6=20+(-2)
18=18
ANSWER: x= -2</span>
