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3 years ago

2.5 + 0.5k=2(k + 0.5)

Mathematics

2 answers:

Alecsey [184]3 years ago

5 0

Answer:

k =1

Step-by-step explanation:

Solve for k:

0.5 k + 2.5 = 2 (k + 0.5)

Expand out terms of the right hand side:

0.5 k + 2.5 = 2 k + 1

Subtract 2 k from both sides:

(0.5 k - 2 k) + 2.5 = (2 k - 2 k) + 1

0.5 k - 2 k = -1.5 k:

-1.5 k + 2.5 = (2 k - 2 k) + 1

2 k - 2 k = 0:

2.5 - 1.5 k = 1

Subtract 2.5` from both sides:

(2.5 - 2.5) - 1.5 k = 1 - 2.5

2.5 - 2.5 = 0:

-1.5 k = 1 - 2.5

1 - 2.5 = -1.5:

-1.5 k = -1.5

Divide both sides of -1.5 k = -1.5 by -1.5:

(-1.5 k)/(-1.5) = (-1.5)/(-1.5)

(-1.5)/(-1.5) = 1:

k = (-1.5)/(-1.5)

(-1.5)/(-1.5) = 1:

Answer: k = 1

sammy [17]3 years ago

5 0

Answer:

k=1

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Answer:

When n is 1

3n+4

=3*1+4

=3+4

When n is 2

3n+4

=3*2+4

=6+4

=10

When n is 3

3n+4

=3*3+4

=9+4

=13

When n is 4

3n +4

=3*4+4

=12+4

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3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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Determine whether the given vectors are orthogonal, parallel, or neither. (a) a = 9, 6 , b = −4, 6 orthogonal parallel neither (

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Answer:

(a) Orthogonal, (b) Neither, (c) Parallel, d( Orthogonal

Step-by-step explanation:

a and b are parallel is a=kb

a and b if a dot b = 0

(a) a dot b = (9)(-4)+6(6)=-36+36=0

a and b are orthogonal

(b) a dot b = 40-7-21=12

a=kb -> there is no k value satisfying the equation

a and b are neither parallel or orthogonal

a=kb -> k=-3/4 satisfies the equation

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(d) a dot b= 9-3-6=0

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