The first thing to do in this case is to find the volume of the sphere.
We have then:
V = (4/3) * (pi) * (r ^ 3)
V = (4/3) * (pi) * ((4) ^ 3)
v = 268 in ^ 3
Then, the but of the sphere will be given by:
W = (0.075) * (268)
W = 20
Answer:
the weight of the sphere, to the nearest pound is:
A) 20
Answer:
875.018
Step-by-step explanation:
To solve this, we must plug in 6 for our t value. So, we have the equation:
-4.982 - 20(6) + 1000
Following the rules of PEMDAS, we have:
-4.982 - 120 + 1000 = 875.018
Answer:
Step 1 : Multiple both sides by 2
3x + 2y = 1
4x - 2y = 20
Step 2 : Eliminate 1 variable by adding the equations
7x = 21
Step 3 : Divide both sides by 7 to solve for x
x = 3
Step 4: Substitute the value of x into 3x + 2y = 1
3(3) + 2y = 1
Step 5: Solve for y
2y = 1 - 9
2y = -8
*divide by 2
y = - 4
Solution:
(x,y) = (3, -4)
<h3>
Answer: 10.1 cm approximately</h3>
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Explanation:
The double tickmarks show that segments DE and EB are the same length.
The diagram shows that DB = 16 cm long
We'll use these facts to find DE
DE+EB = DB
DE+DE = DB
2*DE = DB
DE = DB/2
DE = 16/2
DE = 8
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Now let's focus on triangle DEC. We just found the horizontal leg is 8 units long. The vertical leg is EC which is unknown for now. We'll call it x. The hypotenuse is CD = 9
Use the pythagorean theorem to find x
a^2+b^2 = c^2
8^2+x^2 = 9^2
64+x^2 = 81
x^2 = 81 - 64
x^2 = 17
x = sqrt(17)
That makes EC to be exactly sqrt(17) units long.
If you follow those same steps for triangle ADE, then you'll find the missing length is AE = 6
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So,
AC = AE+EC
AC = 6 + sqrt(17)
AC = 10.1231056256177
AC = 10.1 cm approximately
2/4 is equal to 1/2. Hope this helps
