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Thepotemich [5.8K]
3 years ago
8

Which translation will change figure ABCD to figure A'B'C'D'?

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

5 units left and 7 units up

Step-by-step explanation:

idk  i amn as confused as you

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A linear function is a trend that is equal to each other or corresponds to another set of numbers. The answer here would be C. because the multiples properly align in the table.
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For items 5 - 7 consider the score distribution of 15
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The median is 79. Hope this helps!
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The circumference of the earth is about 25,000 miles. Approximately how far is it from any point on the earth's surface to its c
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Since C = 2rπ
 
if the C of earth = 25000 miles

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3 years ago
If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?
oksano4ka [1.4K]

\boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given \boxed{ \ a(x) = 3x + 1 \ } and \boxed{ \ b(x) = \sqrt{x - 4} \ }.

We will form (b o a)(x) and then determine the domain.

<u>Step-1</u>

\boxed{ \ (b \circ a)(x) = b(a(x)) \ }

Replace each appearance of x in b(x) with \boxed{ \ a(x) = 3x + 1 \ }.

\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }

Thus, \boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }

<u>Step-2</u>

To be defined, the value under the radical sign must not be negative. Therefore, the domain of (b \circ a)(x) = \sqrt{3x - 3} are processed as follows.

\boxed{ \ 3x - 3 \geq 0 \ }

Both sides added by 3.

\boxed{ \ 3x \geq 3 \ }

Both sides divided by 3.

\boxed{ \ x\geq 1 \ }

Thus, the domain of (b \circ a)(x) = \sqrt{3x - 3} is \boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Learn more</h3>
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Keywords: composition of function, if a(x) = 3x + 1, and, b(x) = √(x-4), what is the domain of, (b o a)(x), b(a(x)), defined, the value, under the radical sign, must not be negative,

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3 years ago
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What does h equal in the equation S=2πr2+2πrh?
Troyanec [42]

A

Step-by-step explanation:First, subtract

2

π

r

2

from each side of the equation to isolate the

h

term:

S

−

2

π

r

2

=

2

π

r

h

+

2

π

r

2

−

2

π

r

2

S

−

2

π

r

2

=

2

π

r

h

+

0

S

−

2

π

r

2

=

2

π

r

h

Now, divide each side of the equation by

2

π

r

to solve for

h

:

S

−

2

π

r

2

2

π

r

=

2

π

r

h

2

π

r

S

−

2

π

r

2

2

π

r

=

2

π

r

h

2

π

r

S

−

2

π

r

2

2

π

r

=

h

h

=

S

−

2

π

r

2

2

π

r

Or

h

=

S

2

π

r

−

2

π

r

2

2

π

r

h

=

S

2

π

r

−

2

π

r

2

2

π

r

h

=

S

2

π

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−

r

2

r

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=

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−

r

4 0
3 years ago
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