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3 years ago

I have no clue how to do this please help-

Mathematics

2 answers:

laiz [17]3 years ago

4 0

Answer:

X = 2

Step-by-step explanation:

erica [24]3 years ago

3 0

Answer:

Step-by-step explanation:

(x+1)/2=21/14

cross multiply

14(x+1)=21(2)

14x+14=42

14x=42-14

14x=28

x=2

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Radius:5 cm Diameter: Circumference: Area:

Snowcat [4.5K]

Answer: Diameter = 10cm

Circumference = 31.4cm

Area = 78.57cm2

Step-by-step explanation: Radius = 5cm

→Diameter = 2×radius

= 2×5 = 10cm

Diameter = 10cm

→ Circumference = 2 ×22/7 × 5

= 220/7

=31.4cm

→ Area = 22/7 × 5 × 5

22/7× 25 = 78.57cm2

7 0

3 years ago

A sphere has a radius of 5.3 feet.

Talja [164]

V = (4/3)π*r^3
.. = (4/3)*3.14*(5.3 ft)^3
.. = 623.3 ft^3

5 0

4 years ago

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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

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What did they do wrong

Evgen [1.6K]

Add 4 they were supposed to subtract 4. The subtraction sign is actually represented as a negative 2x. I think..

5 0

3 years ago

What is Three to the power of negative three

Rom4ik [11]

Answer:

0.04

Step-by-step explanation:

7 0

3 years ago

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