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Sergeu [11.5K]
4 years ago
13

Dictionaries are considered a resource allowed in an open book test. Please select the best answer from the choices provided T F

SAT
1 answer:
Katena32 [7]4 years ago
8 0

The answer is true on edg. I just took the test.

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The spontaneous formation of a lipid bilayer in an aqueous environment occurs because:
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I believe the answer is all of the above
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2 years ago
For a lemonade stand, the total cost c, in dollars, of selling n cups of lemonade is given by c=100+ 1.5n. What is the best inte
anygoal [31]

Answer:

OOOOO lemonade sounds good!

Explanation:

4 0
3 years ago
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Which addition expression has the sum 8 – 3i ? (9 2i) (1 – i) (9 4i) (–1 – 7i) (7 2i) (1 – i) (7 4i) (–1 – 7i)
charle [14.2K]

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B- (9 + 4i) + (–1 – 7i)

Explanation:

:)

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3 years ago
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ONLY GODS CAN SOLVE STRAIGHT AWAY: Using the given equation find the missing coordinates of the points and then find the slope o
Yakvenalex [24]

Answer:

By the time Mike left his house at 12:00 noon, Erik had already traveled: 2 x 70 mph =140 miles.

308 - 140 = 168 miles left.

Let the time when they meet =T

70T + 50T =168 miles, solve for T

T =7 / 5 =1.4 hours, or 1 hour and 24 minutes after 12:00 noon.

12 + 1.4 =1 O'clock + 24 minutes, or 24 minutes after 1:00 pm when they will meet.

5 0
3 years ago
Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.
skelet666 [1.2K]

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)

\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4

\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}

6 0
3 years ago
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