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2 years ago

-4.122 as a fraction

Mathematics

1 answer:

miss Akunina [59]2 years ago

4 0

Answer:

-4 61/500

Step-by-step explanation:

just bcuz

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Robin can mow a lawn in 3 hours, while Brady can mow the same lawn in 4 hours. How many hours would it take for them to mow the

BaLLatris [955]

The answer is twelve sevenths (12/7)

x - the job
t - time to do the job x

Robin can mow a lawn in 3 hours: x/3 * t
Brady can mow the same lawn in 4 hours: x/4 * t

If they work together, the total work will be the sum of their speeds:
x = x/3 * t + x/4 * t = (x/3 + x/4)*t = (4x/12 + 3x/12) * t = 7x/12 * t

x = 7x/12 * t

Divide both sides by x:
x/x = 7x/12 * t/x
1 = 7/12 t
t = 1 : 7/12
t = 12/7

5 0

2 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

2 years ago

The lines intersect at point C. What is the value of x? A. 159 B. 100 C. 80 D. 21

KIM [24]

I would say B, but do you have a graph to show? I'm confused.

3 0

3 years ago

Whats the length of the short side?

anzhelika [568]

Call the short side x. We have

tan 72° = opposite/adjacent = (12.8 cm) / x

x = 12.8 / tan 72° = 12.8/(3.01) = 4.158...

Answer: 4.2 cm

3 0

3 years ago

The Capacity of a 1 cm cube is 1 mL. How many 1 cm ice cubes do you need to melt to fill a 1.5 L jar.

Svetach [21]

1500 ice cubes need to fill 1.5 L jar

Solution:

1 cm = 1 ml

Capacity of ice cube = 1 cm

= 1 ml

Capacity of jar = 1.5 L

1 L = 1000 ml

1.5 L = 1.5 × 1000

= 1500 ml

Capacity of jar = 1500 ml

$$\text{Number of ice cubes} =\frac{\text{Capacity of ice cube}}{\text{Capacity of jar}}$

$$=\frac{1500}{1}$

= 1500

Therefore 1500 ice cubes need to fill 1.5 L jar.

8 0

3 years ago