Ask question

Ask question

All categories

jarptica [38.1K]

2 years ago

12

(2x-7)^2 subtracted from 2

1 answer:

weqwewe [10]2 years ago

4 0

Write out the expression 2 - (2x-7)^2. That's it.

But if you want to go further and remove the parentheses, first expand (2x-7)^2: (2x-7)^2 = 4x^2 - 28x + 49,

and then subtract this result from 2:

2 - (4x^2 - 28x + 49) (It's important to use parentheses here)

Now, following the distributive property of multiplication, remove the parentheses:

2 - 4x^2 + 28x - 49

Combining the constants, we get the final answer: - 4x^2 + 28x - 47

You might be interested in

I need the answer! Please help me!

AleksAgata [21]

He needs to work 22 hours :)

6 0

3 years ago

Read 2 more answers

Translate the phrase into an inequality: <br> Seven plus a number x is at most twenty

Len [333]

At most means less than or equal to:

7 + x ≤ 20

3 0

2 years ago

In a class 2/5 of the children are boys.What fraction are girls

Bas_tet [7]

Answer:

3/5

Step-by-step explanation:

Self explanatory.

3 0

1 year ago

Read 2 more answers

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

6 0

2 years ago

Read 2 more answers

Which number line shows all the values of a that make the inequality - 3x +1 < 7 true?

GrogVix [38]

Answer:

Third option

Step-by-step explanation:

5 0

2 years ago