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2 years ago

Find all possible values of b such that 4x² + bx + 3 can be factored. Check all that apply.

Mathematics

1 answer:

aleksklad [387]2 years ago

3 0

The possible values of b are 7 and 8

<h3>How to determine the possible values of x?</h3>

The expression is given as:

4x² + bx + 3

Next, we test the options to determine the values of b

<u>Option 1: b = 13</u>

So, we have:

4x² + 13x + 3 ---- this cannot be factorized

<u>Option 2: b = 7</u>

So, we have:

4x² + 7x + 3

Expand

4x² + 4x + 3x + 3

Factorize

4x(x +1) + 3(x + 1)

Factor out x + 1

(4x + 3)(x + 1) ------ this can be factorized

<u>Option 3: b = 8</u>

So, we have:

4x² + 8x + 3

Expand

4x² + 6x + 2x + 3

Factorize

2x(2x +3) + 1(2x + 3)

Factor out 2x + 3

(2x + 3)(2x + 1) ------ this can be factorized

<u>Option 4: b = 1</u>

So, we have:

4x² + x + 3 ---- this cannot be factorized

Hence, the possible values of b are 7 and 8

Read more about factorized expressions at:

brainly.com/question/723406

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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

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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

$(t) = \ time$

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

$T_{s} = Surrounding \ temperature$

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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

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

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

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

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

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

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