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ale4655 [162]
3 years ago
6

Rectangle abcd is the image of rectangle abcd after a dilation. What is the scale factor of the dilation? Enter your answer in t

he box
Rectangle 1 is a -5, -2
B -8, -2
C-8, -5
D -5, -5
Rectangle 2 is a-2.5,-1
B -4,-1
C-4, -2.5
D-2.5, 2.5
Mathematics
2 answers:
Ivanshal [37]3 years ago
6 0
The scale factor of dilation is 1/2 or
\frac{1}{2}
because the number of rectangle one is multiplied by 1/2 to make rectangle 2


good luck
Vsevolod [243]3 years ago
6 0

Answer:

1/2

Step-by-step explanation:

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andre [41]

The ball describes a parabola, as you can see in the attached picture. So, the point where the ball strike the ground is the point where the parabola meets the x axis. In fact, the x-axis is the set of points where y=0, which means that the ball has height 0 or -again- it hits the ground.

So, we have to set y=0 in our equation and look for the positive solution. We have

-\dfrac{1}{14}x^2+6x+3=0 \iff x=42\pm\sqrt{1806}

And the positive solution is

42+\sqrt{1806}

So that's the distance from the child where the ball strikes the ground.

6 0
3 years ago
For the function f(x)=-3(x-1) find f(0)
pentagon [3]
F(0)=-3(0-1)
f(0)=-3(-1)
f(0)=3
3 0
3 years ago
Which of the following definitions describe functions from the domain to the codomain given? Which functions are one-to-one? Whi
LenKa [72]

Answer:

  • a) f is a function. It is not 1-1, it is not onto.
  • b) g is not a function.
  • c) h is a function. It is not 1-1, it is onto.
  • f) h is a function. It is a bijection, and h^-1(x,y)=(y-1,x-1)

Step-by-step explanation:

a)  For all x∈ℤ, the number f(x)=x²+1 exists and is unique because f(x) is defined using the operations addition (+) and multiplication (·) on ℤ. Then f is a function. f is not one-to-one: consider -1,1∈ℤ. -1≠1 but f(-1)=f(1)=2- Because two different elements in the domain have the same image under f, f is not 1-1. f is not onto: x²≥0 for all x∈ℤ then f(x)=x²+1≥1>0 for all x∈ℤ. Then 0∈ℕ but for all x∈ℤ f(x)≠0, which means that one element of the codomain doen't have a preimage, so f is not onto.

b) 0∈ℕ, so 0 is an element of the domain of g, but g(0)=1/0 is undefined, therefore g is not a function.

c) Let (z,n)∈ℤ x ℕ. The number h(z,n)=z·1/(n+1) is unique and it's always defined because n+1>0, then h is a function. h is not 1-1: consider the ordered pairs (1,2), (2,5). They are different elements of the domain, but h(2,5)=2/6=1/3=h(1,2). h is onto: any rational number q∈ℚ can be written as q=a/b for some integer a and positive integer b. Then (a,b-1)∈ ℤ x ℕ and h(a,b-1)=a/b=q.

f) For all (x,y)∈ℝ², the pair h(x,y)=(y+1,x+1) is defined and is unique, because the definition of y+1 and x+1 uses the addition operation on ℝ. f is 1-1; suppose that (x,y),(u,v)∈ℝ² are elements of the domain such that h(x,y)=h(u,v). Then (y+1,x+1)=(v+1,u+1), so by equality of ordered pairs y+1=v+1 and x+1=u+1. Thus x=u and y=x, therefore (x,y)=(u,v). f is onto; let (a,b)∈ℝ² be an element of the codomain. Then (b-1,a-1)∈ℝ² is an element of the domain an h(b-1,a-1)=(a-1+1,b-1+1)=(a,b). Because h is 1-1 and onto, then h is a bijection so h has a inverse h^-1 such that for all (x,y)∈ℝ² h(h^-1(x,y))=(x,y) and h^-1(h^(x,y))=(x,y). The previous proof of the surjectivity of h (h onto) suggests that we define h^-1(x,y)=(y-1,x-1). This is the inverse, because h(h^-1(x,y))=h(y-1,x-1)=(x,y) and h^-1(h^(x,y))=h^-1(y+1,x+1)=(x,y).

5 0
3 years ago
4y-27=3y then y is equal to ?
guapka [62]

Answer:

y=4/3y -9

Step-by-step explanation:

just divide both sides of the equation by 3

7 0
3 years ago
Five different written driving tests are administered by the Motor Vehicle Department. One of these five tests is selected at ra
kipiarov [429]
<span>P(A) = Number of favorable outcomes to A/ Total number of outcomes 
= 1/5
or 0.2
</span>
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3 years ago
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