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1 year ago

Ratio and Proportion 19. The ratio 9: x is equivalent to 36:20. What is the value of x?

Mathematics

1 answer:

Flauer [41]1 year ago

5 0

Answer:

9:x = 36:20

9/x = 36/20

x/9 =20/36

x=(20/36)×9

x=5

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What is 41 divided by 3605

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A silo is not the shape of a cone. The silo is 8 meters tall and it's base has a diameter of 3 meters. Soybeans cost $414 per cu

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Answer:

$7803.72

Step-by-step explanation:

We have been given that a silo is in the shape of a cone. The silo is 8 meters tall and it's base has a diameter of 3 meters. Soybeans cost $414 per cubic meter. We are asked to find the total cost to fill the silo with soybeans.

First of all, we will find the volume of silo using volume of cone formula.

$\text{Volume of cone}=\frac{1}{3}\pi r^2 h$ , where,

r = radius

h = Height

We know that radius is half the diameter, so radius of silo would be $r=\frac{3}{2}=1.5$ .

$\text{Volume of cone}=\frac{1}{3}\pi (1.5\text{ m})^2 (8)\text{ m}$

$\text{Volume of cone}=\frac{1}{3}\pi\times 2.25\text{ m}^2 (8)\text{ m}$

$\text{Volume of cone}=\pi\times 0.75\text{ m}^2 (8)\text{ m}$

$\text{Volume of cone}=\pi\times 6\text{ m}^3$

$\text{Volume of cone}=18.8495559\text{ m}^3$

Now we will multiply total volume by $414 to find total cost.\

$\text{Total cost of soybeans}=\$414 \times 18.8495559$

$\text{Total cost of soybeans}=\$7803.7161426$

$\text{Total cost of soybeans}\approx \$7803.72$

Therefore, it will cost $7803.72 to fill the silo with soybeans.

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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

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2 years ago

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