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3 years ago

Find the unknown value in the ratio.

Mathematics

1 answer:

Makovka662 [10]3 years ago

4 0

15.75:14=9:8
3:1.875=2:1.25

Hope this helps, although I am not a 100% sure it is correct :)

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Transportation Safety The given expression is used for determining the likelihood that a water taxi will have a total passenger

marysya [2.9K]

Answer:

Value of given expression = 0.4628 (Approx)

Value of given expression [Upto two decimal] = 0.46 (Approx)

Step-by-step explanation:

Missing equation;

[175 - 172] / [29 / √20]

Find:

Value of given expression

Computation:

[175 - 172] / [29 / √20]

[3] / [29 / √20]

[3] / [29 / 4.4721]

[3] / [6.4846]

0.4628

Value of given expression = 0.4628 (Approx)

Value of given expression [Upto two decimal] = 0.46 (Approx)

3 0

3 years ago

What is the model of 5*17

agasfer [191]

5 width 7 legnth gttt gm
bslhskag br Malandrino

7 0

3 years ago

Which fraction represents the product of -5/6 x 5/6 in simplest form? A. -1 B. -5/6 C. -25/36 D. 25/36

barxatty [35]

-25/36 is the correct answer to your question

4 0

3 years ago

Read 2 more answers

How can you identify the different type of solutions to one-variable equations?

Pie

Answer:

An equation like 2x + 3 = 7 is a simple type called a linear equation in one ... There are other types of equations, however, that can have several solutions. ... to solve an equation algebraically and you end up with an obvious identity (like 3 = 3)

Step-by-step explanation:

7 0

4 years ago

When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71. $\overline {6}$ °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

$\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)$

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

$\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min$

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

$\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min$

8 0

3 years ago