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2 years ago

Write the equation 2x + 4y = 12 in slope intercept form. Show your work.

Mathematics

1 answer:

BigorU [14]2 years ago

6 0

Answer:

First we have to find the value of x;

then we have to find the value of y;

Step-by-step explanation:

here,

2x= 12

x = 12\2

x = 6

now,

4y = 12

y = 12\4

y = 3

° the value of x is 6

° ° the value of y is 3

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a 1 kg ball has 10 joules of kinetic energy and starts to roll up a hill how far up the hill will it be when it stops

AURORKA [14]

Answer:

Step-by-step explanation:

A 1 kg ball has 10 joules of kinetic energy and starts to roll up a hill.

As along the hill the ball rises up, it loses its kinetic energy. The kinetic energy is converted to potential energy.

According to the law of conservation of energy, the kinetic energy plus the potential energy equals a constant.

Here given kinetic energy as 10 J, so this energy will get converted to potential energy.

We know that, potential energy is

$P.E=m\cdot g\cdot h$

where,

m is the mass, g is the acceleration due to gravity and h is the height.

Putting the values,

$\Rightarrow 10=1\times 10\times h$

$\Rightarrow 10=10 h$

$\Rightarrow h=1\ m$

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3 years ago

Josh swam 25 laps in 50 minutes. At that rate, how long will it take to swim 35 laps?

GrogVix [38]

Answer:

70 Minutes

Step-by-step explanation:

Divide 25 into 50 to get 2 and then multiply 2 x 35 to get 70

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3 years ago

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

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3 years ago

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Hitman42 [59]

Answer:

Step-by-step explanation:

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Vika [28.1K]

Answer:

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Step-by-step explanation:

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3 years ago

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