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3 years ago

Two numbers are in the ratio of 2 to 3. If the smaller number is 18, the larger number is___.

Mathematics

2 answers:

hodyreva [135]3 years ago

8 0

Answer:

Step-by-step explanation:

Let's set up a proportion. The numbers are in a ratio of 2 to 3, and the smaller number is 18, while the larger number is unknown. We can call it x.

smaller/larger=smaller/larger

2/3=18/x

Cross multiply. Multiply the numerator of the first fraction by the denominator of the second. Then, multiply the numerator of the second by the denominator of the first.

2*x=3*18

2x=54

We want to find out what x is. We have to get x isolated. Currently, x is being multiplied by 2. The inverse of multiplication is division. Divide both sides by 2.

2x/2=54/2

x=54/2

x=27

The larger number is 27.

Tamiku [17]3 years ago

4 0

Answer:

Step-by-step explanation:

2:3

18:?

2 x 9 = 18

3 x 9 = 27

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$\mathrm{ABCD} \sim \mathrm{GFHE}$

Solution:

ABCD and EGFH are two trapezoids.

To determine the correct way to tell the two trapezoids are similar.

Option A: $\mathrm{ABCD} \sim \mathrm{GFHE}$

AB = GF (side)

BC = FH (side)

CD = HE (side)

DA = EG (side)

So, $\mathrm{ABCD} \sim \mathrm{GFHE}$ is the correct way to complete the statement.

Option B: $\mathrm{ABCD} \sim \mathrm{EGFH}$

In the given image length of AB ≠ EG.

So, $\mathrm{ABCD} \sim \mathrm{EGFH}$ is the not the correct way to complete the statement.

Option C: $\mathrm{ABCD} \sim \mathrm{FHEG}$

In the given image length of AB ≠ FH.

So, $\mathrm{ABCD} \sim \mathrm{FHEG}$ is the not the correct way to complete the statement.

Option D: $\mathrm{ABCD} \sim \mathrm{HEGF}$

In the given image length of AB ≠ HE.

So, $\mathrm{ABCD} \sim \mathrm{HEGF}$ is the not the correct way to complete the statement.

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A chemical flows into a storage tank at a rate of (180+3t) liters per minute, where t is the time in minutes and 0<=t<=60

Yuliya22 [10]

Answer:

The amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

Step-by-step explanation:

Consider the provided information.

A chemical flows into a storage tank at a rate of (180+3t) liters per minute,

Let $c(t)$ is the amount of chemical in the take at <em>t </em>time.

Now find the rate of change of chemical flow during the first 20 minutes.

$\int\limits^{20}_{0} {c'(t)} \, dt =\int\limits^{20}_0 {(180+3t)} \, dt$

$\int\limits^{20}_{0} {c'(t)} \, dt =\left[180t+\dfrac{3}{2}t^2\right]^{20}_0$

$\int\limits^{20}_{0} {c'(t)} \, dt =3600+600$

$\int\limits^{20}_{0} {c'(t)} \, dt =4200$

So, the amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

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