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3 years ago

What is the solution to 3(2x-1)=(6x+5)

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

3 0

We are given this equation and I'm assuming you want us to solve for x

$3(2x-1)=6x+5$

Use the distributive property on the left side of the equation

$6x-3=6x+5$

Subtract both sides by 6x

$-3=5$

After trying to solve for x, we have this weird equation. Since -3 does NOT equal 5 and there is no other way for us to try and solve this, the equation has no solution.

Let me know if you need any clarifications, thanks!

~ Padoru

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55 is what percent of 100

max2010maxim [7]

55 is 55% of 100.

percents are based of of 100, so 55 would be 55/100 or 55%.

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3 years ago

the height of the snow was 84 inches in the beginning of the week. it melts at a rate of 0.15 inches per hour. how many hours wo

lakkis [162]

Answer:

It takes 147 hours for the snow to reach 62 inches.

Step-by-step explanation:

Given:

Height of the snow at the beginning = 84 inches

Rate at which the snow melts = f 0.15 inches per hour

To Find:

The time taken for the snow to reach 62 inches = ?

Solution:

Let the time taken to reach 62 inches be x

Then according to the question,

Total height of the snow - (rate of melting)(x) = 62 inches

On substituting the values,

84 - (0.15)(x) = 62

84 - 0.15x = 62

- 0.15x = 62 - 84

- 0.15x = -22

0.15x = 22

x = \frac{22}{0.15}

x= 146.6

x = 147

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3 years ago

Graph the quadratic function f(x)= -2x^2 + 6x - 2

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The answer is........

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3 years ago

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Complete the table and write the equation for the relationship shown in the graph.

SVEN [57.7K]

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3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago