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2 years ago

12

D. 360 DCW Help please !!! My answer is C not Sure

1 answer:

tino4ka555 [31]2 years ago

7 0

Flipping something 180 degrees (doesn't matter if clockwise or counterclockwise) will always flip that thing upside down.

Your answer is 180 degrees.

Now, the reason why it's not 270 degrees clockwise because that's pretty much turning that 90 degrees counterclockwise. And 90 degrees is about how much a car does when making a left or right turn.

Have an awesome day! :)

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How is constructing an angle bisector similar to constructing a perpendicular bisector?

Sophie [7]

Hello,
Your answer would be D.

4 0

2 years ago

If f(x) = 2x ^ 2 - 5 and g(x) = x ^ 2 - 4x - 8 find (f - g)(x)

choli [55]

Answer: (f-g)(x) is equivalent to f(x) - g(x)

So that would be (2x²-5) - (x²-4x-8) = 2x² - x² + 4x + 8 - 5 = x² + 4x + 3

Hope this helps.

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2 years ago

Help me on number 10 please.

LenaWriter [7]

Answer:

9.) B. division

10.) 8

Explanation:

9. Which operation to do first in order of operations?

Answer: Applying BODMAS where:

[B] brackets

[O] order

[D] division

[M] multiplication

[A] addition

[S] subtraction

Maintaining order, division operation is done first. Then multiplication and then addition and so on.

10. Given that N = 48 ÷ 3 × 2

⇒ N = 48 ÷ 3 × 2

⇒ N = 48 ÷ 6

⇒ N = 8

3 0

11 months ago

Read 2 more answers

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago

Algebra 1 ty brainilest if correct

exis [7]

Answer:

1 exponential decay and 78%

5 0

2 years ago

Read 2 more answers