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3 years ago

Help please Rewrite as an improper fraction.

Mathematics

2 answers:

NARA [144]3 years ago

4 0

The answer is the second option, 37/11.

3/4/11 = 3 wholes of 11 = 33.
add the remaining 4 to 33 and u get 37. thus answer is 37/11

kozerog [31]3 years ago

3 0

Answer:

The answer is option b) 37/11

Step-by-step explanation:

3 4/11 = ( (11 × 3) + 4)11

= 37/11

The answer is option b) 37/11

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The product of two consecutive numbers is 4 more than three times the larger. find the integer

Stella [2.4K]

Answer:

Step-by-step explanation:

Let the first integer be x, then the next is x+1.

x(x + 1) = 3(x + 1) + 4

x^2 + x = 3x+ 3 + 4

x^2 + x - 3x - 7 = 0

x^2 - 2x - 7 = 0

Solving this does not give an integer so maybe there's something wrong with the question??

7 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

3 years ago

-3 (3c + 5) = 2 (10 - c)

Ilya [14]

Answer:-5

Step-by-step explanation:

You need to distribute on both sides so u get

9c-15=20-2c

Move variable to left side and change the sign

-9c+2c-15=20

Move constant to right side and change the sign

-9c+2c=20+15

Connect like terms

-7c=35

Divide

C=-5

5 0

2 years ago

Find f(2) if f(x) = (x 1)2.

Katyanochek1 [597]

I hope this helps you

x=2

f (2)=(2+1)^2=3^2=9

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Download photo math and it will let you take a picture of the equation and answer it for you

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3 years ago

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