Observe the graph below. The scale factor used to dilate ABC was...

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

7 0

Answer:

Step-by-step explanation:

AB = 1 unit

A'B' = 3 units

The scale of dilation = A'B'/AB

Therefore:

scale of dilation = 3 units / 1 unit = 3

This means that ABC was dilated or enlarged in size by multiplying every of its segment by 3. In essence, A'B'C' is 3 times the size of ABC.

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Help needed. here is some text

slamgirl [31]

Checking the graph:- when x = 1 y = 0 so it cant be y = x^4

y = log4 x
is the same as

x = 4^y

Checking the graph, when x = 4, y = 1
Plugging these into the above 4 = 4^1 = 4

So the correct choice is first one

8 0

3 years ago

Solve h = −16t2 + 36t + 4.

kvasek [131]

I can do each of the things I asked above.
So, let's change this into vertex form:
$h=-16t^2+36t+4$
$h=(-16t^2+36t)+4$
$h=-16(t^2-2.25t)+4$
$h=-16(t^2-2.25t+1.27-1.27)+4$
$h=-16(t^2-2.25t+1.27)+20.32+4$
$h=-16(t-1.125)^2+24.32$
The vertex is at (1.125,24.32)
Answers may vary due to rounding

Factored Form:
$h = -16t^2 + 36t + 4$
$h = -4\left(4t^2-9t-1\right)$

Quadratic Formula:
$x = \frac{-b +/- \sqrt{b^2-4(a)(c)}}{2a}$
h = -16t^2 + 36t + 4
a = -16 b = 36 c = 4
$h = \frac{-(36) +/- \sqrt{(36)^2-4(-16)(4)}}{2(-16)}$
$h = \frac{-36 +/- \sqrt{1552}}{-32}$
$h = ≈-0.11$
$h = ≈2.36$

5 0

2 years ago

Help me please.No spammmers.

egoroff_w [7]

Answer:

Step-by-step explanation:

cot =cos/sin =a/b

cos=a/b*sin

(sin-cos)/(sin+cos) =(sin-sin*a/b)(sin+sin*a/b)

=(1-a/b)/(a+a/b)

=(b-a)/(a+b)

5 0