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3 years ago

Estimate the quotient: 1628 ÷ 8

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

6 0

Answer:

Answer is: 203.5

Step-by-step explanation:

Hope this helped YOU! :)

Pachacha [2.7K]3 years ago

4 0

Answer:

203.5

Step-by-step explanation:

Please mark brainliest.

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Which of the following equations corresponds to the graph below?

mrs_skeptik [129]

Answer: D: Y= -2x - 1

Step-by-step explanation:

y= mx+b

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3 years ago

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On a particular day, the wind added 2 miles per hour to Jaime's rate when she was rowing with the wind and subtracted 2 miles pe

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Take 42-2 = 40 and 34+2 = 36 so Jaime's normal rowing speed was 36 miles

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3 years ago

The difference of this equation.

yaroslaw [1]

Answer:

$\frac{23}{(y-8)(y+8)}=\frac{23}{y^{2}-64}$

Step-by-step explanation:

$\frac{3y - 1}{(y-8)(y+8)} -\frac{3}{y+8} = \frac{3y - 1}{(y-8)(y+8)} -\frac{3(y-8)}{(y-8)(y+8)} =\\\\=\frac{3y - 1-3y+24}{(y-8)(y+8)} = \frac{23}{(y-8)(y+8)}=\frac{23}{y^{2}-64}$

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3 years ago

YEAR 9 MATHS- PLEASE HELP

Lilit [14]

Answer:

y=2

Step-by-step explanation:

since the line is horizontal, on the point 2 on the y axis, the equation is 2. If, however the line were to be diagonal, it would have been different, as would it if the line were vertical(in which case, the equated would start x =)

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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

3 years ago