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4 years ago

A spherical paintball measures 1.5 centimeters in diameter. Approximately how many paint is in it?

1 answer:

8 0

For this case, what we must do is find the volume of the sphere, which will be given by:
V = (4/3) * (pi) * (r ^ 3)
Where,
r = radius of the sphere
Substituting the values we have:
V = (4/3) * (pi) * ((1.5 / 2) ^ 3)
V = 1.767145868
Round:
V = 1.78 cm ^ 3
Answer:
There are approximately 1.78 cm ^ 3 of paint.

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Please help me answer I need this question right, one more wrong answer and I'll fail

Shkiper50 [21]

Answer: The second option.

Step-by-step explanation:

1. Calculate the first volume assuming that the figure is completed (with no missing part). Therefore, you must apply the formula for calculate the volume of a square prism:

$V_A=s^{2}*h$

Where s is the side of the base and h is the height.

2. Then:

$V_A=(40cm)^{2}*50cm=80,000cm^{3}$

4. The volume of the missing part is calculated with the formula of a rectangular prism:

$V_B=l*w*h$

Where l is the lenght, w is the width and h is the height. Then:

$V_B=(40cm)(40cm-35cm)(50cm-25cm)=5,000cm^{3}$

5. Therefore, you must subtract both volumes to obtained the volume of the figure:

$V_f=80,000cm^{3}-5,000cm^{3}=75,000cm^{3}$

6 0

3 years ago

Multiply. −2 1/4⋅(−2 1/3) A −5 1/4 B −4 1/12 C 4 1/12 D 5 1/4

vodka [1.7K]

The answer is D
You multiply the denominator with the whole number and and the numerator

4 0

3 years ago

Read 2 more answers

Refer to the accompanying data set and construct a 9090% confidence interval estimate of the mean pulse rate of adult females;

Lilit [14]

Answer:

B i think

Step-by-step explanation:

5 0

3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Write the equation (in point-slope form) for the line parallel to y = x + 4 and passes

Tanya [424]

Answer:

$y = x-7$

Step-by-step explanation:

Step 1: Write the equation in point-slope form

Point Slope Form: $y-y_{1}=m\left(x-x_{1}\right)$

Plug in the numbers and solve

$y - (-2) = 1(x - 5)$

$y + 2 = x - 5$

$y + 2 - 2 = x - 5 - 2$

$y = x-7$

Answer: $y = x-7$

8 0

3 years ago

Read 2 more answers