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3 years ago

2. Find (w2 + 1)(w3 - W+1).

Mathematics

1 answer:

12345 [234]3 years ago

4 0

Answer:

$w^{5} + w^{2} - w + 1$

Step-by-step explanation:

We have to find the simplified expression of $(w^{2} + 1) (w^{3} - w + 1)$ .

Now, we have to multiply each term using the distributive property of multiplication.

So, $(w^{2} + 1) (w^{3} - w + 1)$

= $w^{2} \times (w^{3} - w + 1) + 1 \times (w^{3} - w + 1)$

= $(w^{5} - w^{3} + w^{2} ) + (w^{3} - w + 1)$

= $w^{5} + w^{2} - w + 1$ (Answer)

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What is 8,792 as a fraction?

Wittaler [7]

We have a number $8.792$ .

To write it as a fraction first count the number of decimals and call it $n$ (which, in our case, has a value of 3). Multiply the number with $10^n$ to get $8792$ . Then divide this number by $10^n$ to get $\frac{8792}{10^n}$ from here you can either simplify (if the assignment says so, then the simplified fraction is $\frac{1099}{125}$ ) or not.

Hope this helps.

6 0

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

6 0

2 years ago

Read 2 more answers

WILL MARK BRAINLIEST<br><br>Which equation can be used to find the volume of this solid?<br>

AURORKA [14]

V=(7*6)/2 *8 the third option

you multiply the height with the base of the triangle and you get the area of the base of the prism so you multiply it with the height of the prism to get the volume

5 0

3 years ago

Just answer number 7 and 8

melamori03 [73]

We are given the equation:
F = 2.25+0.2(m-1)
where:
F is the fare
m is the number of miles.

In question 7, we are given that the fare (F) is equal to $6.05 and we need to get the number of miles. To do so, we will simply substitute with the fare in the given equation and solve for the number of miles (m) as follows:
F = 2.25 + 0.2(m-1)
6.05 = 2.25 + 0.2(m-1)
6.05-2.25 = 0.2(m-1)
3.8 = 0.2(m-1)
3.8/0.2 = m-1
19 = m-1
m = 19+1
m = 20 miles

Number 8 is exactly the same, but we will substitute F=7.65 and again solve for m

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3 years ago

Which of these equations would give you the balance on an account with a principal of $3000 and an interest rate of 8%, compound

GrogVix [38]

It would be 3000 x 1.08^5

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3 years ago