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kenny6666 [7]
2 years ago
8

A box holds three circular flower pots, each with the same diameter. What is the exact area of the base of one pot? Show your wo

rk.( the length of the box is 15 in.)
Mathematics
1 answer:
RoseWind [281]2 years ago
7 0
19.635 square inches.

Assuming that the pots are arranged in a straight line, each pot would have a diameter of 5, because 5 • 3 = 15, and 15 is the length of the box. The radius of the base of one pot is therefore 2.5.
A = pi • r^2 = pi • 2.5^2 = 19.635 (approximation)
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dmitriy555 [2]
Use formula y2-y1 over x2 - x1

3-0/4-0

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3 years ago
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X, Y and Z form the vertices of a triangle. XY = 12.4m, XZ = 10.4m and YZ = 8.7m. Find the angle ∠ YXZ rounded to 1 DP.
sweet-ann [11.9K]

Answer:

43.8°

Step-by-step explanation:

Applying,

Cosine rule,

From the diagram attached,

x² = y²+z²-2yxcos∅.................... Equation 1

where ∅ = ∠YXZ

Given: x = 8.7 m, y = 10.4 m, z = 12.4 m

Substitute these values  into equation 1

8.7² = 10.4²+12.4²-[2×10.4×12.4cos∅]

75.69 = (108.16+153.76)-(257.92cos∅)

75.69 = 261.92-257.92cos∅

collect like terms

257.92cos∅ = 261.92-75.69

257.92cos∅ = 186.23

Divide both sides by the coefficient of cos∅

cos∅ = 186.23/257.92

cos∅ = 0.722

Find the cos⁻¹ of both side.

∅ = cos⁻¹(0.7220)

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∅ = 43.8°

6 0
3 years ago
an engineer decides to use similar triangles to find the distance across the river she makes the diagram below which triangles t
sergiy2304 [10]
<h2>Explanation:</h2><h2 />

The missing diagram is shown below. Remember that shapes are similar if you can turn one into the other by moving, rotating, flipping, or scaling. So this means you can make the object bigger or smaller. So in this case, we know that ΔABC is similar to ΔDEF. If so, we can establish the following proportion:

\frac{h}{56.7}=\frac{17.2}{52.4} \\ \\ h=56.7\times \frac{17.2}{52.4} \\ \\ h=18.61m

Our goal is to find x. So by Pythagorean theorem:

x=\sqrt{17.2^2+18.61^2} \\ \\ \boxed{x=25.34m}

Conclusion:

<em>ΔABC ~ΔDEF</em>

<em>Distance across the river is 25.34m </em>

8 0
3 years ago
How do you solve #13
joja [24]
I have no idea sorry i figure it out brb
6 0
3 years ago
At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
beks73 [17]
<h2>Answer:</h2>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

<h2>Step-by-step explanation:</h2>

In the question,

Taking the elevation of pool along the y-axis, and length of the board along the x-axis.

On drawing the illustration in the co-ordinate system we get,

lₓ = 2 m

uₓ = 2.5 m/s

and,

h_{y}=9.5\,m

So,

From the equations of the laws of motion we can state that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}

So,

On putting the values we can say that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=(0)t+\frac{1}{2}(9.8)t^{2}\\t^{2}=\frac{9.5}{4.9}\\t^{2}=1.93\\t=1.39\,s

Now,

The <u>equation of the motion in the horizontal</u> can be given by,

v_{x}=u_{x}+a_{x}t\\v_{x}=2.5+a_{x}(1.39)\\So,\\v_{x}=1.39a_{x}+2.5

<em><u>Therefore, the equations of the motions in the horizontal and verticals are,</u></em>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

6 0
3 years ago
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