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uranmaximum [27]
3 years ago
6

Are these answers correct? If not, what would the answers be?

Mathematics
1 answer:
Ludmilka [50]3 years ago
7 0
Yes you have the correct answers for each of the four problems. Nice work.

For anyone curious,
rd = removable discontinuity
id = infinite discontinuity
c = continuous
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I need to simplify this expression : 6• (10•k)
Temka [501]

Answer:

6 x 10= 60

60/2=30

Answer is<u> 30</u> i think

Step-by-step explanation:

7 0
4 years ago
I don't know how to do this but I know you have to find the growth or decay
xenn [34]

Answer:

\boxed{y=5(1.05)^t\to\:5\% \:rate \:of \:growth}

\boxed{f(t)=50(.95)^t\to\:5\% \:rate \:of \:decay}

\boxed{g(t)=50(1.5)^t\to\:50\% \:rate \:of \:growth}

\boxed{y=5(.5)^t\to\:50\% \:rate \:of \:decay}

Step-by-step explanation:

We can rewrite the given functions in the form f(x)=a(100\%+r\%)^x to determine the rate of growth or f(x)=a(100\%-r\%)^x to determine the rate of decay

y=5(1.05)^t=5(100\%+5\%)^t\to\:5\% \:rate \:of \:growth

f(t)=50(.95)^t=50(100\%-5\%)^t\to\:5\% \:rate \:of \:decay

g(t)=50(1.5)^t=50(100\%+50\%)^t\to\:50\% \:rate \:of \:growth

y=5(.5)^t=5(100\%-50\%)^t\to\:50\% \:rate \:of \:decay

4 0
4 years ago
A 3 by 3cm grid shows nine 1 by 1 cm squares and uses 24cm of wire.
zalisa [80]

Answer:

840 cm

Step-by-step explanation:

Given that in the question we have, : 9, 1 cm × 1 cm squares setting a 3 × 3 grid.

Therefore wire required in 3 x 3 grid is

3(3+1) + 3(3+1)

= 3 × 4 + 3 × 4

= 24 I'm

Similarly, the length of the wire required in 20 by 20 grid will be

=> 20(20+1) + 20(20+1)

=> 20 × 21 + 20 × 21

=> 420 + 420

= 840cm

6 0
3 years ago
PLEASE ANSWER ALL THE QUESTIONS WILL MARK BRAINLIEST!!!
Doss [256]

Answer:

1. A

2. C

3. A

Explanation:

5x+(-2)=6x+4

-5x -5x

(-2)=1x+4

-6=x

3 0
3 years ago
Trigonometric identities
Goryan [66]

Without knowing what Juan's exact steps were, it's hard to say what he did wrong. The least you could say is that his solution is simply not correct.

4 sin²(<em>θ</em>) - 1 = 0

==>   sin²(<em>θ</em>) = 1/4

==>   sin(<em>θ</em>) = ±1/√2

==>   <em>θ</em> = <em>π</em>/4, 3<em>π</em>/4, 5<em>π</em>/4, 7<em>π</em>/4

7 0
3 years ago
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