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Oliga [24]
3 years ago
15

What is the distance between the points (5.-18) and (8,

Mathematics
2 answers:
romanna [79]3 years ago
8 0

Answer:

\sqrt{10}

Step-by-step explanation:

The horizontal distance from points (5,-18) and (8,-17) is 3 because it is 3 units from 5 to 8.  The vertical distance is 1 since it is one unit from -18 to -17.  Now we can use the equation a^{2}+b^{2}=c^{2} where a=3 and b=1 and c is the distance that you are looking for:

a^{2} +b^{2} =c^{2}\\3^{2} +1^{2} =c^{2}\\9+1=c^{2}\\10=c^{2}\\\sqrt{10}=c

Ivenika [448]3 years ago
6 0

3.1622776601684

Step-by-step explanation:

X=8-5=3

Y=-17 - -18=1

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What are the coordinates of the vertices of the pre-image given? ry= −x ◦ T1, −2(x, y) 
nataly862011 [7]
<h3><u>Answer:</u></h3>

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

<h3><u>Step-by-step explanation:</u></h3>

We have to find the  coordinates of the vertices of the pre-image given?

Ry= −x ◦ T_1, −2(x, y)

i.e. we have to find the composition of reflection along the line y=-x and translation with the rule:

(x,y) → (x+1,y-2)

Now we have coordinates of A",B",C" and D" as:

A"(-4,-2)

B"(-2,-1)

C"(-2,-4)

D"(-4,-3)

Now we are asked to find the coordinates of the pre-image.

Also when any point is reflected along y=-x then the point is transformed to:

(x,y) → (-y,-x)

Let A,B,C and D are the points of the pre-image.

Hence, the coordinates of the pre-image are given as:

so, the transformation is given as:

A→A'→A"

B→B'→B"

C→C'→C"

D→D'→D"

Where A',B',C',D' represents the transformation after translation and A",B",C",D" represents the transformation after reflection as well.

The coordinates of A' are (2,4)

B' are (1,2)

C' are (4,2)

and D' are (3,4)

Now, the coordinates of pre-image are given as:

A'(x,y)  → A(x-1,y+2)=A(1,6)

B'(x,y)  → B(x-1,y+2)=B(0,4)

C'(x,y)  → C(x-1,y+2)=C(3,4)

and D'(x,y)  → D(x-1,y+2)=D(2,6)

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

4 0
3 years ago
given two terms in a geometric sequence find the 8th term and the recursive formula . a4=-12 and a5=-6
Crazy boy [7]

A geometric sequence is defined by a starting point, a, and a common ratio r

The first term is a, and you get every next term by multiplying the previous one by r.

So, our terms are

\left[\begin{array}{c|c}a_1&a\\a_2&ar\\a_3&ar^2\\a_4&ar^3=-12\\a_5&ar^4=-6\end{array}\right]

We can see that when we pass from a_4 to a_5 the number gets halved (-12 \mapsto -6)

This implies that the common ratio is r = \frac{1}{2}

So, the table becomes

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\dfrac{1}{8}a = -12 \iff a = -12\cdot 8 = -96

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a_n = -\dfrac{96}{2^{n-1}}

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AysviL [449]

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