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2 years ago

What are the factors of 32Round the following items to the nearest £1:

Mathematics

1 answer:

Nonamiya [84]2 years ago

5 0

The factors of 32 as required in the task content are; 2, 4, 8, 16, 32.

<h3>What are the factors of 32?</h3>

The factors of the number 32 as required in the task content are numbers whose multiples amount to 32 and are as listed above.

The item prices as given in the task content rounded to the nearest £1 are;

Notebook

£2.15 = £2

Pencil Case

£3.80 = £4

Colouring Pencil

£1.49 = £1

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Determine whether the graph of y^5=x^3 is symmetric with respect to the Y axis the x-axis or the origin.

zhannawk [14.2K]

9514 1404 393

Answer:

the origin

Step-by-step explanation:

Both terms are of odd degree, so the function is an odd function. It is symmetrical with respect to the origin.

If you recast this as y = f(x), then ...

f(x) = x^(3/5)

When we look at f(-x), we find ...

f(-x) = ((-x)^3)^(1/5) = -(x^3)^(1/5) = -(x^(3/5)) = -f(x)

An odd root of a number has the same sign as the number.

When f(-x) = -f(x), the function is odd and symmetrical about the origin.

7 0

3 years ago

Translate the phrase into a numerical expression. One hundred devided by the quantity three times five

tester [92]

Answer:

$\dfrac{100}{3 \times 5}$

Step-by-step explanation:

$\dfrac{100}{3 \times 5}$

6 0

2 years ago

PLEASE ANSWER FAST AS POSSIBLE I HAVE 40 LESSONS DUE BY 12 PM :(

Sophie [7]

Try B, hope that helps :)

5 0

3 years ago

Read 2 more answers

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.

5 0

3 years ago

Read 2 more answers

At a theme park, 35% of daily visitors ride the log ride. If on one particular day the park has 60,000 visitors, how many can be

Neko [114]

The answers is D 21,000because you have to do 60,000x.35=21,000

3 0

3 years ago

Read 2 more answers