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-BARSIC- [3]
3 years ago
12

Rectangle ABCD has vertices A(8, 5) , B(8, 10) , C(14, 10) , and D(14, 5) . A dilation with a scale factor of 1.2 and centered a

t the origin is applied to the rectangle. Which vertex in the dilated image has coordinates of (9.6, 6) ?
A′

B′

C′

D′
Mathematics
2 answers:
chubhunter [2.5K]3 years ago
8 0

Answer:

The answer is A'

Step-by-step explanation:

I did this one before

love history [14]3 years ago
4 0
Rectangle ABCD has vertices A(8, 5) , B(8, 10) , C(14, 10) , and D(14, 5) . A dilation with a scale factor of 1.2 and centered at the origin is applied to the rectangle. Which vertex in the dilated image has coordinates of (9.6, 6) ?

A′ (8×1.2=9.6,5×1.2=6)

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If two tangent segments to a circle share a common endpoint outside a circle, then the two segments are
matrenka [14]

If two tangent segments to a circle share a common endpoint outside a circle, then the two segments are congruent. This is according to the intersection of two tangent theorem. The theorem states that  given a circle, if X is any point within outside the circle and if Y and Z are points such that XY and XZ are tangents to the circle, then XY is equal to XZ.

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4 0
4 years ago
A rectangle has a height of x+3 and a width of x+7. what is the area?
stiks02 [169]

Answer:

4x+7

Step-by-step explanation:

6 0
4 years ago
When solving the system x + 2y = 9 and x - y = 6 by substitution, what can be substituted for x in the second equation?
bazaltina [42]
X+2y = 9 ---------eq.1
x-y =6 ----------- eq.2

x = 9 -2y (from eq. 1)

Now substitute the value of x in the second eq.2

9-2y-y =6
-3y = -3
y =1

So,
x +2 (1)= 9
x= 7


8 0
4 years ago
Log5(x-1)+log5(x+3)-1=0
jonny [76]

Answer:

there are no real solutions

Step-by-step explanation:

\log_5(x-1)+\log_5(x+3)-1=0\\\\\log_5(x-1)+\log_5(x+3)=1

there is a rule that says

\log_b(a)+\log_b(c)=\log_b(ac)

so we have

\log_5(x-1)+\log_5(x+3)=\log_5((x-1)(x+3))=1

and we have the definition

\log_a(b)=c\\\\a^{c} =b

so we have

5^{1} = (x-1)(x+3)\\\\5=x^{2} -1x+3x-3\\\\5=x^{2} +2x-3\\\\0=x^{2} +2x-3-5\\\\0=x^{2} +2x-8\\\\

and using the quadratic formula we get that

there are no real solutions

3 0
4 years ago
Read 2 more answers
Factorise b^2 + 7b + 10
deff fn [24]

Step-by-step explanation:

b²+7b+10

Common factors are 5,2

b²+7b+10

b²+2b+5b+10

b(b+2)+5(b+2)

(b+2)(b+5)

Hope it helps.✨✨

4 0
2 years ago
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