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3 years ago

How many solutions does the equation have?

Mathematics

2 answers:

lesya692 [45]3 years ago

7 0

I hope this helps you

Elena L [17]3 years ago

3 0

Answer:

option C (many)

Step-by-step explanation:

$4(z-5)+2=4z-18$

Distribute the numbers inside the parenthesis

$4(z-5)+2=4z-18$

$4z-20+2=4z-18$

Combine like terms

Subtract 4x from both sides

$4z-20+2=4z-18$

$-20+2=-18$

$-18=-18$

Both sides are equal. So there are more than one solution for x

Answer is option C (many)

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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

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

Solve with interval notation Solve: 5 | x − 2 | + 4 > 8

Gnesinka [82]

Answer:

x<6/5, x>14/5

Step-by-step explanation:

Steps

$5\left|x-2\right|+4>8$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$5\left|x-2\right|+4-4>8-4$

$\mathrm{Simplify}$

$5\left|x-2\right|>4$

$\mathrm{Divide\:both\:sides\:by\:}5$

$\frac{5\left|x-2\right|}{5}>\frac{4}{5}$

$\mathrm{Simplify}$

$\left|x-2\right|>\frac{4}{5}$

$\mathrm{Apply\:absolute\:rule}:\quad\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad\mathrm{or}\quad\:u\:>\:a$

$x-2<-\frac{4}{5}\quad\mathrm{or}\quad\:x-2>\frac{4}{5}$

Show Steps

$x-2<-\frac{4}{5}\quad:\quad x<\frac{6}{5}$

Show Steps

$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$

$\mathrm{Combine\:the\:intervals}$

$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$

5 0

3 years ago

A football field is 300 ft x 160 ft . If A coach ask his players to run from one corner to the opposite corner diagonally, how f

Elan Coil [88]

48000f2t2 pretty sure that is it

7 0

3 years ago

Read 2 more answers

Find all of the zeros of the function:

IceJOKER [234]

Answer:Roots: 3;2;-1

Step-by-step explanation: i use horner's scheme for approximating the roots of polynomials

6 0

3 years ago

The table gives the surface areas, in square kilometres, of five seas.

zlopas [31]

The table is missing, so i have attached it.

Also, the correct question is;

The surface area of the East China Sea is k times the surface area of the Baltic Sea.

Work out the value of k.

Give your answer to the nearest whole number

Answer:

k = 3

Step-by-step explanation:

From the attached table,

Surface area of east china = 1.25 × 10^(6) km²

Surface area of Baltic sea = 4.22 × 10^(5) km² = 0.422 × 10^(6) km²

We are told that the surface area of East China Sea is k times that of the surface area of the Baltic Sea.

Thus;

1.25 × 10^(6) = k × 0.422 × 10^(6)

k = 1.25/0.422

k = 2.96

Approximating to whole number gives;

k = 3

4 0

3 years ago