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3 years ago

What is the absolute value of 1/2?

1 answer:

7 0

The Absolute Value of a number is "the magnitude of a real number without regard to its sign". So the Absoulute Value of 1/2 is |1/2|

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I AM GIVING AWAY 25 POINTS TO BEST ANSWER PLEASE HELP! I AM GIVING LOTS OF POINTS FOR THIS! Cube shaped boxes of tissue are ship

Vika [28.1K]

Try finding the area of the box and the container
Then divide the container's area by the tissue box

Hope this helps :)

4 0

3 years ago

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A ride ticket usually costs

weqwewe [10]

Answer: Hope this helps if this is what you want.

$14.25

Step-by-step explanation:

$1.50 = 1 ticket

1.50 * 10 = 15

$15 = 10 tickets

100% - 5% = 95% (0.95)

15 * 0.95 = 14.25

$14.25

7 0

1 year ago

Of the 400 employees, 5% of them are park managers. How many managers does the park employe

Anvisha [2.4K]

Answer:

The park employed 20 managers.

Step-by-step explanation:

5% of 400

(5 ÷ 100) × 400

0.05 × 400

= 20

8 0

3 years ago

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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

2 years ago

The table shows the heights of students in a group

8_murik_8 [283]

Answer:

Mean=(46+48+51+53)/4

Mean=198/4

Mean=49.5

Step-by-step explanation:

Mean is the total amount per the total number of terms. It is also known as average.

8 0

2 years ago

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