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Usimov [2.4K]
3 years ago
8

Ivan and Tanya share £150 in the ratio 4: 1 Work out how much more Ivan gets compared to Tanya.

Mathematics
2 answers:
dalvyx [7]3 years ago
7 0

Answer:

£90 more

Step-by-step explanation:

sum the parts of the ratio, 4 + 1 = 5 parts

Divide the amount by 5 to find the value of one part of the ratio

£150 ÷ 5 = £30 ← value of 1 part of the ratio, then

4 parts = 4 × £30 = £120 ← Ivan's share

£120 - £30 = £90

Ivan got £90 more than Tanya

nalin [4]3 years ago
4 0

Answer:

Ivan gets 90 more

Step-by-step explanation:

Divide 150 by 5 to get tanyas share, 30 pound, then subtract 30 from the 120 ivan gets

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Work out the nth of this sequence:
Gekata [30.6K]

Answer:

a_n = -11 +3(n -1)

Explanation:

The nth term of an arithmetic sequence is explicitly defined as a_n = a_1 +d(n -1) where a_1 is the first term of the sequence and d is the the common difference.

From the given first five terms of the sequence we can see that the first term is 11 so a_1 = 11.

The common difference, d, can be calculated by a_n - a_{n -1} so we'll find the common difference of the given sequence by letting n = 2

d = a_2 - a_{2 -1} \\d = a_2 -a_{1} \\d = -8 -(-11) \\ d = -8 +11 \\ d = 3.

Now let's plug everything we know.

a_1 = -11

d = 3

a_n = -11 +3(n -1)

6 0
2 years ago
If joan reads 96 pages in 4 hours how long will it take her to read 240
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Answer:

10 hours to readm240

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
3 years ago
-1,2,5,7...with these numbers while using multiplaction, divison, addition, and subtraction, how would u get positive 24
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By adding two positives together, or a big positive number with a small negative number.
4 0
3 years ago
1. A football team gains 2 yards on the first play,
Ilya [14]

Option C

The football team had a overall loss of 2 yards

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When the team gains yards we use a positive value, and when the team loses yards we use a negative value.

<em><u>Given that, football team gains 2 yards on the first play</u></em>

First play = +2

<em><u>Given that football team loses 5 yards on the second play</u></em>

Second play = -5

<em><u>Given that football team loses 3 yards  on the third play</u></em>

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Let us simplify

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So, -2 represents loss of two yards (since negative value indicates loss)

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