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3 years ago

Work out the balance for £240 invested for 20 years at 7% per annum

Mathematics

1 answer:

Luda [366]3 years ago

5 0

Step-by-step explanation:

It depends in the type on interest

Simple Interest

PxRxT

£240 x 0.07x 3 = £50.4

£240 + £50.4 = £290.4

Compound Interest

Px(1+R)^T

£290.4 x (1 + 0.07)³ = £355.7524872= €355.76

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It’s triangular prism

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2 years ago

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An astronomical unit (AU) is the average distance between Earth and the sun. One AU is equal to approximately 150,000,000 km. As

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2 years ago

The mayor of a town has proposed a plan for the annexation of an adjoining community. A political study took a sample of 1600 vo

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Answer:

Test statistic = -2.44

There is enough evidence to support the strategist's claim.

Step-by-step explanation:

H0 : p = 0.41

H1 : p < 0.41

pˆ = 0.38

Test statistic :

z=pˆ−p/√p(1−p)/n

Z = (0.38 - 0.41) / √(0.41(1 - 0.41) / 1600

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2 years ago

Unit activity: exponential and logarithmic functions

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We will conclude that:

The domain of the exponential function is equal to the range of the logarithmic function.
The domain of the logarithmic function is equal to the range of the exponential function.

<h3>Comparing the domains and ranges.</h3>

Let's study the two functions.

The exponential function is given by:

f(x) = A*e^x

You can input any value of x in that function, so the domain is the set of all real numbers. And the value of x can't change the sign of the function, so, for example, if A is positive, the range will be:

y > 0.

For the logarithmic function we have:

g(x) = A*ln(x).

As you may know, only positive values can be used as arguments for the logarithmic function, while we know that:

$\lim_{x \to \infty} ln(x) = \infty \\\\ \lim_{x \to0} ln(x) = -\infty$

So the range of the logarithmic function is the set of all real numbers.

<h3>So what we can conclude?</h3>

The domain of the exponential function is equal to the range of the logarithmic function.
The domain of the logarithmic function is equal to the range of the exponential function.

If you want to learn more about domains and ranges, you can read:

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2 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

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3 years ago

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