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sergeinik
13 days ago
10

The distance, d, in inches of a weight attached to a spring from its equilibrium as a function of time, t, in seconds can be mod

eled by the graph below. Which equation is represented in the graph below? On a coordinate plane, a curve crosses the y-axis at (0, negative 5). It increases to (1, 5) and then decreases to (2, negative 5). 5 cycles are shown.
Mathematics
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A periscope is 5 feet above the surface of the ocean. Through it can be seen a ship that rises to 50 feet above the water. To th
PIT_PIT [12445]

The maximum distance visible on Earth is calculated using the formula d=\sqrt{8,000h_1} + \sqrt{8,000h_2}

h_1= your initial height and h_2= your secondary height

In this case, h_1=5 represents the height of the periscope and h_2=50 denotes the height of the ship, leading us to a distance of 832.45553203368 feet. However, rounding to the nearest mile yields the answer as 11^a^0

If there are any discrepancies, please inform me and I will recalculate!

5 0
2 months ago
The graph of f(x) = x2 is translated to form g(x) = (x – 5)2 + 1. Which graph represents g(x)?
Leona [12618]
<span>The graph will shift 5 units to the right and 1 unit upwards, forming a parabola that opens up with its vertex positioned at (5, 1).

Explanation:
The subtraction of 5 from x prior to squaring indicates a horizontal movement of 5 units to the right.

The addition of 1 signifies a vertical shift of 1 unit up.

This transformation follows the vertex form of a parabola, y=a(x-h)^2 + k, where (h, k) represents the vertex. In this case, h is 5 and k is 1, placing the vertex at (5, 1).</span>
6 0
2 months ago
Read 2 more answers
The Homerun Hitter’s Academy charges thirty-five dollars per hour for batting lessons, with a registration fee of fifteen dollar
AnnZ [12381]

Response:

the result is 6789

Detailed breakdown:

hope this assists❤️

can I get brainliest?❤️

3 0
1 month ago
Read 2 more answers
Using the U- Substitution u=sqrt(2x), integral form 2-8 dx/ sqrt(2x) + 1 is equivalent to ...
AnnZ [12381]
Let's apply u-substitution:u= \sqrt{2x}, \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u}In the substitution process:dx=u du. then the integral transforms to:\int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx=u-ln(u+1)=\sqrt{2x}-ln( \sqrt{2x}+1). Next, we will adjust the limits: \sqrt{16}-ln( \sqrt{16}+1)-( \sqrt{4}-ln( \sqrt{4}+1))=4-ln(5)-2+ln(3)=2+ln(0.6)=2-0.51=1.49

8 0
1 month ago
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