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3 years ago

If x+3/3 = y+2/2 , then x/3=?

Mathematics

1 answer:

igomit [66]3 years ago

6 0

Answer:

y/3

Step-by-step explanation:

x+1=y+1

x=y+1-1

x=y

Therefore, x/3=y/3

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Johnny went out to dinner with his two friends. Johnnys meałeast $12.32, Marks

IrinaVladis [17]

Answer:

Step-by-step explanation:

8 0

3 years ago

2(x)+2(2)=2(x/2)-2(x)<br> Solve this please and show the steps!!!!!!

Olin [163]

Answer: $x=-\frac{4}{3}$

Step-by-step explanation:

$2(x)+2(2)=2(\frac{x}{2})-2(x)$

Remove all those parentheses by multiplying. Note: 2 and 2 cancel out.

$2x+4=x-2x$

Subtract 2x

$4=x-2x-2x$

Combine like terms;

$4=-3x$

Divide by -3

$\frac{4}{-3}=x$

Rewriting it;

$x=-\frac{4}{3}$

5 0

3 years ago

Read 2 more answers

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

3x + 2y = 14. X + y = 5. Work put the values of x and y

andrey2020 [161]

Answer:

x = 4 , y = 1

Step-by-step explanation:

3x + 2y = 14 → (1)

x + y = 5 → (2)

Multiplying (2) by - 2 and adding to (1) will eliminate the y- term

- 2x - 2y = - 10 → (3)

Add (1) and (3) term by term to eliminate y

x = 4

Substitute x = 4 into (2) and evaluate for y

4 + y = 5 ( subtract 4 from both sides )

y = 1

7 0

2 years ago

A researcher believes that 5% of pet dogs in Europe are Labradors. If the researcher is right, what is the probability that the

DerKrebs [107]

Answer:

0.9036

Step-by-step explanation:

Calculation to determine the probability that the proportion of Labradors

P(Proportion greater than 4%)

= P(z> 0.04 -0.05 /√0.05 * 0.95/806

= P(z > -1.30)

=0.9036

Thereforethe probability that the proportion of Labradors is =0.9036

6 0

3 years ago